Re: [algogeeks] os question

2012-12-10 Thread sahil gupta
It's b.
Windows follow this Operation.

On Fri, Dec 7, 2012 at 4:21 AM, manish narayan.shiv...@gmail.com wrote:

  Four processes of 1gb,1.2gb,2gb,2gb are there and RAM available is 2gb.
 We have a
 time shared system. Which of the following is the most appropriate
 scheduling algorithm?
 a. all processes are loaded sequentially 1 by 1
 b. load one process at a time and execute processes in RR fashion
 c. load 1gb, 1,2gb first then processes 3 and 4 follow
 d. All processes can be loaded together and CPU time shared among them

 --




-- 




Re: [algogeeks] os question

2012-12-10 Thread Navin Kumar
If virtualization is concerned, then answer would be choice d. Since its
not necessary to load complete process in memory.


On Sat, Dec 8, 2012 at 12:45 AM, sahil gupta sahilgupta...@gmail.comwrote:

 It's b.
 Windows follow this Operation.


 On Fri, Dec 7, 2012 at 4:21 AM, manish narayan.shiv...@gmail.com wrote:

  Four processes of 1gb,1.2gb,2gb,2gb are there and RAM available is 2gb.
 We have a
 time shared system. Which of the following is the most appropriate
 scheduling algorithm?
 a. all processes are loaded sequentially 1 by 1
 b. load one process at a time and execute processes in RR fashion
 c. load 1gb, 1,2gb first then processes 3 and 4 follow
 d. All processes can be loaded together and CPU time shared among them

 --




  --




-- 




[algogeeks] os question

2012-12-06 Thread manish
 Four processes of 1gb,1.2gb,2gb,2gb are there and RAM available is 2gb. We 
have a
time shared system. Which of the following is the most appropriate 
scheduling algorithm?
a. all processes are loaded sequentially 1 by 1
b. load one process at a time and execute processes in RR fashion
c. load 1gb, 1,2gb first then processes 3 and 4 follow
d. All processes can be loaded together and CPU time shared among them

-- 




[algogeeks] OS and Networking group

2012-11-22 Thread V K Pandey
Hi All,

  Please suggest any one, group or forums related to Operating System 
and Network  like algogeeks. 

Thanks
Vivek P 
 

-- 




[algogeeks] OS question..

2012-11-04 Thread manish
Q1.  If we have infinite memory, then do we still be needing paging?
Q2. Given only 8bits registers, you have to find average of 4 bit registers 
values without using any operation involving 16 bit calculations.

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Re: [algogeeks] OS question..

2012-11-04 Thread Hanlei Qin
I think the answer to Q1 may Yes.
Cause the virtual memory of program is limited, they need logically
contiguous memory, and have limit from OS and processor(32-bit, or
64-bit) yet.
I have no idea about Q2.


On Mon, Nov 5, 2012 at 4:30 AM, manish narayan.shiv...@gmail.com wrote:
 Q1.  If we have infinite memory, then do we still be needing paging?
 Q2. Given only 8bits registers, you have to find average of 4 bit registers
 values without using any operation involving 16 bit calculations.

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[algogeeks] OS question

2012-01-25 Thread UTKARSH SRIVASTAV
I have a doubt when each process has it's own separate page table then why
is there s system wide page table required ? Also if Page table is such
that it maps virtual address to a physical address then I think two process
may map to same physical address because all process have same virtual
address space

any good link on system wide page table may also solve my problem

-- 
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CSE-3
B-Tech 3rd Year
@MNNIT ALLAHABAD*

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Re: [algogeeks] OS question

2012-01-25 Thread Rahul
Google search this
6.033
You will get the basics of processor mode of execution
and rings of execution
Hope I got the question !
On Wed, Jan 25, 2012 at 4:21 PM, UTKARSH SRIVASTAV
usrivastav...@gmail.comwrote:

 I have a doubt when each process has it's own separate page table then why
 is there s system wide page table required ? Also if Page table is such
 that it maps virtual address to a physical address then I think two process
 may map to same physical address because all process have same virtual
 address space

 any good link on system wide page table may also solve my problem

 --
 *UTKARSH SRIVASTAV
 CSE-3
 B-Tech 3rd Year
 @MNNIT ALLAHABAD*


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Re: [algogeeks] OS question

2012-01-25 Thread UTKARSH SRIVASTAV
can't get :(

On Wed, Jan 25, 2012 at 5:19 PM, Rahul raikra...@gmail.com wrote:

 Google search this
 6.033
 You will get the basics of processor mode of execution
 and rings of execution
 Hope I got the question !
 On Wed, Jan 25, 2012 at 4:21 PM, UTKARSH SRIVASTAV 
 usrivastav...@gmail.com wrote:

 I have a doubt when each process has it's own separate page table then
 why is there s system wide page table required ? Also if Page table is such
 that it maps virtual address to a physical address then I think two process
 may map to same physical address because all process have same virtual
 address space

 any good link on system wide page table may also solve my problem


 --
 *UTKARSH SRIVASTAV
 CSE-3
 B-Tech 3rd Year
 @MNNIT ALLAHABAD*


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@MNNIT ALLAHABAD*

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[algogeeks] os ques about paging

2011-09-24 Thread sivaviknesh s
Consider a virtual memory system in which the virtual page addresses are
mapped onto physical page addresses as follow
Virtual page address.. Physical page address
03
12
2...1
The address of any byte in this system is given by the ordered pair (v,d)
where v = virtual page address d = byte offset in the page
Assuming that the main memory is capable of holding 4 page frames, what is
the physical address of a byte having the virtual address (0,512) given that
the page size is 1024 bytes?
a) 3584
b) 2560
c) 1536
d) 512
e) 1024


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Re: [algogeeks] os ques about paging

2011-09-24 Thread Vishnu Ganth
3584

On Sun, Sep 25, 2011 at 9:53 AM, sivaviknesh s sivavikne...@gmail.comwrote:




 Consider a virtual memory system in which the virtual page addresses are
 mapped onto physical page addresses as follow
 Virtual page address.. Physical page address
 03
 12
 2...1
 The address of any byte in this system is given by the ordered pair (v,d)
 where v = virtual page address d = byte offset in the page
 Assuming that the main memory is capable of holding 4 page frames, what is
 the physical address of a byte having the virtual address (0,512) given that
 the page size is 1024 bytes?
 a) 3584
 b) 2560
 c) 1536
 d) 512
 e) 1024


 --
 Regards,
 $iva

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Re: [algogeeks] os ques about paging

2011-09-24 Thread aditya kumar
c) 512

On Sun, Sep 25, 2011 at 9:53 AM, sivaviknesh s sivavikne...@gmail.comwrote:




 Consider a virtual memory system in which the virtual page addresses are
 mapped onto physical page addresses as follow
 Virtual page address.. Physical page address
 03
 12
 2...1
 The address of any byte in this system is given by the ordered pair (v,d)
 where v = virtual page address d = byte offset in the page
 Assuming that the main memory is capable of holding 4 page frames, what is
 the physical address of a byte having the virtual address (0,512) given that
 the page size is 1024 bytes?
 a) 3584
 b) 2560
 c) 1536
 d) 512
 e) 1024


 --
 Regards,
 $iva

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Re: [algogeeks] os ques about paging

2011-09-24 Thread aditya kumar
m sry i dint read the question properly
page no = vitual page % 3 ie 0 % 3 = 3
thrfre 3*1024 is starting addr of physical page
but the byte address will be 3*1024 + 512 = 3584

On Sun, Sep 25, 2011 at 10:12 AM, Vishnu Ganth crazyvishnu...@gmail.comwrote:

 3584


 On Sun, Sep 25, 2011 at 9:53 AM, sivaviknesh s sivavikne...@gmail.comwrote:




 Consider a virtual memory system in which the virtual page addresses are
 mapped onto physical page addresses as follow
 Virtual page address.. Physical page address
 03
 12
 2...1
 The address of any byte in this system is given by the ordered pair (v,d)
 where v = virtual page address d = byte offset in the page
 Assuming that the main memory is capable of holding 4 page frames, what is
 the physical address of a byte having the virtual address (0,512) given that
 the page size is 1024 bytes?
 a) 3584
 b) 2560
 c) 1536
 d) 512
 e) 1024


 --
 Regards,
 $iva

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Re: [algogeeks] OS

2011-09-14 Thread Anup Ghatage
Address Space:
The total addresses taken up by a process is known as a process's address
space

On Wed, Sep 14, 2011 at 7:48 PM, teja bala pawanjalsa.t...@gmail.comwrote:

 can any one tell the difference between  ADDRESS SPACE and VIRTUAL
 ADDRESS SPACE?

 thx in advance.

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Re: [algogeeks] OS

2011-09-14 Thread mohan kumar
http://www.tenouk.com/WinVirtualAddressSpace.html

go to this link.
think it will help you...
i

On Wed, Sep 14, 2011 at 7:48 PM, teja bala pawanjalsa.t...@gmail.comwrote:

 can any one tell the difference between  ADDRESS SPACE and VIRTUAL
 ADDRESS SPACE?

 thx in advance.

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Re: [algogeeks] os

2011-09-08 Thread sharmila saru
4

On Wed, Sep 7, 2011 at 9:09 PM, Mohit Goel mohitgoel291...@gmail.comwrote:

 How many processes are created in this snippet?
 Main()
 {
 Fork();
 Fork()  fork () || fork ();
 Fork ();
 }

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[algogeeks] os

2011-09-07 Thread Mohit Goel
How many processes are created in this snippet?
Main()
{
Fork();
Fork()  fork () || fork ();
Fork ();
}

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Re: [algogeeks] os

2011-09-07 Thread vivek goel
hey is it  4 child processes along wid one parent
process.??
m i rite??

On Wed, Sep 7, 2011 at 9:09 PM, Mohit Goel mohitgoel291...@gmail.comwrote:

 How many processes are created in this snippet?
 Main()
 {
 Fork();
 Fork()  fork () || fork ();
 Fork ();
 }

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Re: [algogeeks] os

2011-09-07 Thread Mohit Goel
a. 15
b. 19
c. 21
d. 27
e. 31
these are the only options.

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Re: [algogeeks] os

2011-09-07 Thread rahul vatsa
19 prs will be created, total 20 prs.
 we have discussed this a few days back. plz check the old thread for any
explanation.

On Wed, Sep 7, 2011 at 12:01 PM, Mohit Goel mohitgoel291...@gmail.comwrote:

 a. 15
 b. 19
 c. 21
 d. 27
 e. 31
 these are the only options.

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Re: [algogeeks] os

2011-09-07 Thread vivek goel
can anyone explain me how?? plsss

On Wed, Sep 7, 2011 at 9:40 PM, rahul vatsa vatsa.ra...@gmail.com wrote:

 19 prs will be created, total 20 prs.
  we have discussed this a few days back. plz check the old thread for any
 explanation.


 On Wed, Sep 7, 2011 at 12:01 PM, Mohit Goel mohitgoel291...@gmail.comwrote:

 a. 15
 b. 19
 c. 21
 d. 27
 e. 31
 these are the only options.

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Re: [algogeeks] os

2011-09-07 Thread Dheeraj Sharma
wats the ans... 21??

On Wed, Sep 7, 2011 at 9:42 PM, vivek goel vivek.thapar2...@gmail.comwrote:

 can anyone explain me how?? plsss


 On Wed, Sep 7, 2011 at 9:40 PM, rahul vatsa vatsa.ra...@gmail.com wrote:

 19 prs will be created, total 20 prs.
  we have discussed this a few days back. plz check the old thread for any
 explanation.


 On Wed, Sep 7, 2011 at 12:01 PM, Mohit Goel mohitgoel291...@gmail.comwrote:

 a. 15
 b. 19
 c. 21
 d. 27
 e. 31
 these are the only options.

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Re: [algogeeks] os

2011-09-07 Thread Dheeraj Sharma
ups...19 i guess

On Wed, Sep 7, 2011 at 9:52 PM, Dheeraj Sharma
dheerajsharma1...@gmail.comwrote:

 wats the ans... 21??


 On Wed, Sep 7, 2011 at 9:42 PM, vivek goel vivek.thapar2...@gmail.comwrote:

 can anyone explain me how?? plsss


 On Wed, Sep 7, 2011 at 9:40 PM, rahul vatsa vatsa.ra...@gmail.comwrote:

 19 prs will be created, total 20 prs.
  we have discussed this a few days back. plz check the old thread for any
 explanation.


 On Wed, Sep 7, 2011 at 12:01 PM, Mohit Goel 
 mohitgoel291...@gmail.comwrote:

 a. 15
 b. 19
 c. 21
 d. 27
 e. 31
 these are the only options.

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Re: [algogeeks] os

2011-09-07 Thread rahul vatsa
int main()
{
   fork();
   fork()  fork() || fork();
   fork();

   return 0;
}


ln 1 : will create 2 prs
ln 2 : will create 10 process for each existing pr
ln 3 : will do fork for all 10 process, nd so now u ve 20 prs

the main issue is @ln-2 in main( )

Ln 2 :

here for 1st fork, if its parent, its ret value in nonzero, so it will
proceed for next fork, as that is for .  if ret value of this 2nd fork is
nonzero, it willn't try the 3rd fork (as its nt required, bcoz that comes
after ||, nd the st is already a true st) , else 2 prs.  --   So for parent
of 1st fork in the st, total no of prs created is 3.

if 1st fork returns 0, ie if it is child, it willn't attempt the next fork(bcoz
thatz for operator  , and the st is already 0), bt it will do attempt the
3rd fork(bcoz that is for operator ||)  so total no processes created here
is - 2


Ln 3- this fork is called for all created processes.  2(Ln 1) * 5(ln 2) * 2
(ln 3) == 20 prs





On Wed, Sep 7, 2011 at 12:12 PM, vivek goel vivek.thapar2...@gmail.comwrote:

 can anyone explain me how?? plsss


 On Wed, Sep 7, 2011 at 9:40 PM, rahul vatsa vatsa.ra...@gmail.com wrote:

 19 prs will be created, total 20 prs.
  we have discussed this a few days back. plz check the old thread for any
 explanation.


 On Wed, Sep 7, 2011 at 12:01 PM, Mohit Goel mohitgoel291...@gmail.comwrote:

 a. 15
 b. 19
 c. 21
 d. 27
 e. 31
 these are the only options.

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Re: [algogeeks] os

2011-09-07 Thread Mohit Goel
20 is not in option ..so whats the answer??

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Re: [algogeeks] os

2011-09-07 Thread Kamakshii Aggarwal
@mohit:answer is 19.one is the parent process originally.and 19 more
processes have been created.

On Wed, Sep 7, 2011 at 11:21 PM, Mohit Goel mohitgoel291...@gmail.comwrote:

 20 is not in option ..so whats the answer??

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Re: [algogeeks] os

2011-09-07 Thread Rashmi Jain
its 19...as 19 prcses are created

On Wed, Sep 7, 2011 at 11:21 PM, Mohit Goel mohitgoel291...@gmail.comwrote:

 20 is not in option ..so whats the answer??

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Re: [algogeeks] os

2011-09-07 Thread Mohit Goel
thnks everyone...

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Re: [algogeeks] os

2011-08-14 Thread aalam roy
@ankur i think you are talking about cleanup handlers. these are the
functions which are executed when a thread terminates. but can you give any
hint how it can be accomplished using process control block.

On Sun, Aug 14, 2011 at 2:15 PM, Kamakshii Aggarwal
kamakshi...@gmail.comwrote:

 @ankur::nahi samajh aaya..:(


 On Sun, Aug 14, 2011 at 9:46 AM, Ankur Khurana 
 ankur.kkhur...@gmail.comwrote:

 My 2 cents,
 When the termination signal is sent to the thread either synchronously or
 asynchronously , you just have a mechanism in place that if that thread is
 in critical section , it exits from there and and unlocks the mutex at point
 of exit. This can be  done by  associating a tokken with the thread that it
 is executing in critical section and what mutex it locked. O we can do this
 by making some table in Process control block.

 On Sat, Aug 13, 2011 at 7:56 PM, Kamakshii Aggarwal 
 kamakshi...@gmail.com wrote:


 How do you make sure to unlock a mutex which was locked in a thread that
 dies/terminates?
 --
 Regards,
 Kamakshi
 kamakshi...@gmail.com

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Re: [algogeeks] os

2011-08-14 Thread Ankur Khurana
@roy : yes , kind of , i didnt know the exact technical term for it. Not
exactly PCB but process can maintain a lookup table for all the shared
variable and there corresposing threads . or for every thread the shared
variable. or ,may be in the thread itself, you can have a linked list
pointer which points to all the shared variable list. Before terminating ,
get that address. I mean it is imagination, i suggested some ways but i
exactly dont know what Windows or Unix employs to do the task.

On Sun, Aug 14, 2011 at 3:03 PM, aalam roy aalamr...@gmail.com wrote:

 @ankur i think you are talking about cleanup handlers. these are the
 functions which are executed when a thread terminates. but can you give any
 hint how it can be accomplished using process control block.


 On Sun, Aug 14, 2011 at 2:15 PM, Kamakshii Aggarwal kamakshi...@gmail.com
  wrote:

 @ankur::nahi samajh aaya..:(


 On Sun, Aug 14, 2011 at 9:46 AM, Ankur Khurana 
 ankur.kkhur...@gmail.comwrote:

 My 2 cents,
 When the termination signal is sent to the thread either synchronously or
 asynchronously , you just have a mechanism in place that if that thread is
 in critical section , it exits from there and and unlocks the mutex at point
 of exit. This can be  done by  associating a tokken with the thread that it
 is executing in critical section and what mutex it locked. O we can do this
 by making some table in Process control block.

 On Sat, Aug 13, 2011 at 7:56 PM, Kamakshii Aggarwal 
 kamakshi...@gmail.com wrote:


 How do you make sure to unlock a mutex which was locked in a thread that
 dies/terminates?
 --
 Regards,
 Kamakshi
 kamakshi...@gmail.com

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[algogeeks] os

2011-08-13 Thread Kamakshii Aggarwal
How do you make sure to unlock a mutex which was locked in a thread that
dies/terminates?
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Re: [algogeeks] os

2011-08-13 Thread sagar pareek
You can make a routine check for mutex

On Sat, Aug 13, 2011 at 7:56 PM, Kamakshii Aggarwal
kamakshi...@gmail.comwrote:


 How do you make sure to unlock a mutex which was locked in a thread that
 dies/terminates?
 --
 Regards,
 Kamakshi
 kamakshi...@gmail.com

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Re: [algogeeks] os

2011-08-13 Thread Ankur Khurana
My 2 cents,
When the termination signal is sent to the thread either synchronously or
asynchronously , you just have a mechanism in place that if that thread is
in critical section , it exits from there and and unlocks the mutex at point
of exit. This can be  done by  associating a tokken with the thread that it
is executing in critical section and what mutex it locked. O we can do this
by making some table in Process control block.

On Sat, Aug 13, 2011 at 7:56 PM, Kamakshii Aggarwal
kamakshi...@gmail.comwrote:


 How do you make sure to unlock a mutex which was locked in a thread that
 dies/terminates?
 --
 Regards,
 Kamakshi
 kamakshi...@gmail.com

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Re: [algogeeks] os

2011-08-11 Thread rajeev bharshetty
Round Robin ..

On Thu, Aug 11, 2011 at 11:01 AM, siddharam suresh
siddharam@gmail.comwrote:

 shortest preemptive  job first
 Thank you,
 Siddharam



 On Thu, Aug 11, 2011 at 10:49 AM, krishna meena 
 krishna.meena...@gmail.com wrote:

 Consider a set of n teaks with known runtimes  r1,r2,r3rn to be
 run on a uni-processor machine. which processor scheduling algorithm
 will result in the maximum throughput?

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[algogeeks] Os/processor dependencies of object file(C compiled file)

2011-08-10 Thread mithun bs
Hi,

I know that the compiled code of a C file(after assembler converts assembly
code to opcode) cannot be run on a different OS or it cannot be run on a
different processor architecture.

So, I need to know what are the machine dependencies which are added in
object file.

One thing is the opcode will be different for each processor architecture.
But how is it dependent on Operating system?


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[algogeeks] os

2011-08-10 Thread krishna meena
Consider a set of n teaks with known runtimes  r1,r2,r3rn to be
run on a uni-processor machine. which processor scheduling algorithm
will result in the maximum throughput?

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Re: [algogeeks] os

2011-08-10 Thread siddharam suresh
shortest preemptive  job first
Thank you,
Siddharam


On Thu, Aug 11, 2011 at 10:49 AM, krishna meena
krishna.meena...@gmail.comwrote:

 Consider a set of n teaks with known runtimes  r1,r2,r3rn to be
 run on a uni-processor machine. which processor scheduling algorithm
 will result in the maximum throughput?

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Re: [algogeeks] os

2011-08-09 Thread raghavendhra rahul
Shared memory is fastest IPC mechanism, since it doesn’t involve any system
call as it is done in user space.
-- 

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Re: [algogeeks] os

2011-08-09 Thread Varun Jakhoria
@Raghvendhra +1 ... because it doesn't require entry at kernel

On Tue, Aug 9, 2011 at 10:55 PM, raghavendhra rahul
rahulraghavend...@gmail.com wrote:
 Shared memory is fastest IPC mechanism, since it doesn’t involve any system
 call as it is done in user space.
 --

 Regards
 Raghavendhra



 changing the face can change nothing .. but facing the change can change
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Re: [algogeeks] os

2011-08-09 Thread *$*
shared memory is fastest IPC mechanism , because , it is a simple memory
allocation on physical memory ,
in case of other options like pipes etc , they requires kernel entries ..

Thx,
--Gopi

On Tue, Aug 9, 2011 at 11:00 PM, Varun Jakhoria varunjakho...@gmail.comwrote:

 @Raghvendhra +1 ... because it doesn't require entry at kernel

 On Tue, Aug 9, 2011 at 10:55 PM, raghavendhra rahul
 rahulraghavend...@gmail.com wrote:
  Shared memory is fastest IPC mechanism, since it doesn’t involve any
 system
  call as it is done in user space.
  --
 
  Regards
  Raghavendhra
 
 
 
  changing the face can change nothing .. but facing the change can
 change
  everything
 
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[algogeeks] os

2011-08-08 Thread Kamakshii Aggarwal
Fastest IPC mechanism is

   1.   ?shared memory
   2.   ?pipes
   3.   ?named pipes
   4.   ?Semaphores

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Re: [algogeeks] os

2011-08-08 Thread sachin sharma
Shared Memory...

On Mon, Aug 8, 2011 at 5:36 PM, Kamakshii Aggarwal kamakshi...@gmail.comwrote:


 Fastest IPC mechanism is

1.   ?shared memory
2.   ?pipes
3.   ?named pipes
4.   ?Semaphores

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Re: [algogeeks] os

2011-08-08 Thread Himanshu Srivastava
named pipes!!!

On Mon, Aug 8, 2011 at 5:36 PM, Kamakshii Aggarwal kamakshi...@gmail.comwrote:


 Fastest IPC mechanism is

1.   ?shared memory
2.   ?pipes
3.   ?named pipes
4.   ?Semaphores

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 Kamakshi
 kamakshi...@gmail.com

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Re: [algogeeks] os

2011-08-08 Thread sachin sharma
http://en.wikipedia.org/wiki/Shared_memory

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Re: [algogeeks] os

2011-08-08 Thread Aditya Virmani
shared memory

On Mon, Aug 8, 2011 at 5:43 PM, Himanshu Srivastava 
himanshusri...@gmail.com wrote:

 named pipes!!!

 On Mon, Aug 8, 2011 at 5:36 PM, Kamakshii Aggarwal 
 kamakshi...@gmail.comwrote:


 Fastest IPC mechanism is

1.   ?shared memory
2.   ?pipes
3.   ?named pipes
4.   ?Semaphores

 --
 Regards,
 Kamakshi
 kamakshi...@gmail.com

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Re: [algogeeks] os

2011-08-08 Thread Himanshu Srivastava
*shared memory is the fastest IPC mechanism
Because we need not copy some data from one place to another.*

On Mon, Aug 8, 2011 at 5:36 PM, Kamakshii Aggarwal kamakshi...@gmail.comwrote:


 Fastest IPC mechanism is

1.   ?shared memory
2.   ?pipes
3.   ?named pipes
4.   ?Semaphores

 --
 Regards,
 Kamakshi
 kamakshi...@gmail.com

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Re: [algogeeks] os

2011-08-08 Thread dilip makwana
Yes NAMED PIPES ... is correct

On 8 August 2011 17:43, Himanshu Srivastava himanshusri...@gmail.comwrote:

 named pipes!!!

 On Mon, Aug 8, 2011 at 5:36 PM, Kamakshii Aggarwal 
 kamakshi...@gmail.comwrote:


 Fastest IPC mechanism is

1.   ?shared memory
2.   ?pipes
3.   ?named pipes
4.   ?Semaphores

 --
 Regards,
 Kamakshi
 kamakshi...@gmail.com

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Re: [algogeeks] os

2011-08-08 Thread Kamakshii Aggarwal
According to me,it should be shared memory..but i have taken this from an
online test which say that the answer is named pipeswhat are named
pipes?
On Mon, Aug 8, 2011 at 5:50 PM, dilip makwana dilipmakwa...@gmail.comwrote:

 Yes NAMED PIPES ... is correct


 On 8 August 2011 17:43, Himanshu Srivastava himanshusri...@gmail.comwrote:

 named pipes!!!

 On Mon, Aug 8, 2011 at 5:36 PM, Kamakshii Aggarwal kamakshi...@gmail.com
  wrote:


 Fastest IPC mechanism is

1.   ?shared memory
2.   ?pipes
3.   ?named pipes
4.   ?Semaphores

 --
 Regards,
 Kamakshi
 kamakshi...@gmail.com

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Re: [algogeeks] os

2011-08-08 Thread dilip makwana
Thanx for correction ... :D

On 8 August 2011 17:50, dilip makwana dilipmakwa...@gmail.com wrote:

 Yes NAMED PIPES ... is correct


 On 8 August 2011 17:43, Himanshu Srivastava himanshusri...@gmail.comwrote:

 named pipes!!!

 On Mon, Aug 8, 2011 at 5:36 PM, Kamakshii Aggarwal kamakshi...@gmail.com
  wrote:


 Fastest IPC mechanism is

1.   ?shared memory
2.   ?pipes
3.   ?named pipes
4.   ?Semaphores

 --
 Regards,
 Kamakshi
 kamakshi...@gmail.com

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 BTech Computers Engineering
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Re: [algogeeks] os

2011-08-08 Thread Gaurav Menghani
I would rather not discuss non-algorithm questions on this group :)

On Mon, Aug 8, 2011 at 6:02 PM, dilip makwana dilipmakwa...@gmail.com wrote:
 Thanx for correction ... :D

 On 8 August 2011 17:50, dilip makwana dilipmakwa...@gmail.com wrote:

 Yes NAMED PIPES ... is correct

 On 8 August 2011 17:43, Himanshu Srivastava himanshusri...@gmail.com
 wrote:

 named pipes!!!

 On Mon, Aug 8, 2011 at 5:36 PM, Kamakshii Aggarwal
 kamakshi...@gmail.com wrote:

 Fastest IPC mechanism is

   ?shared memory
   ?pipes
   ?named pipes
   ?Semaphores

 --
 Regards,
 Kamakshi
 kamakshi...@gmail.com

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Re: [algogeeks] os

2011-08-08 Thread sagar pareek
Named pipes are just like pipes which is global for every process and each
one can access them
so u can say that named pipes are shared global pipes
and i think they are fastest.
pipes works in queue fashion

On Mon, Aug 8, 2011 at 6:13 PM, Gaurav Menghani
gaurav.mengh...@gmail.comwrote:

 I would rather not discuss non-algorithm questions on this group :)

 On Mon, Aug 8, 2011 at 6:02 PM, dilip makwana dilipmakwa...@gmail.com
 wrote:
  Thanx for correction ... :D
 
  On 8 August 2011 17:50, dilip makwana dilipmakwa...@gmail.com wrote:
 
  Yes NAMED PIPES ... is correct
 
  On 8 August 2011 17:43, Himanshu Srivastava himanshusri...@gmail.com
  wrote:
 
  named pipes!!!
 
  On Mon, Aug 8, 2011 at 5:36 PM, Kamakshii Aggarwal
  kamakshi...@gmail.com wrote:
 
  Fastest IPC mechanism is
 
?shared memory
?pipes
?named pipes
?Semaphores
 
  --
  Regards,
  Kamakshi
  kamakshi...@gmail.com
 
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[algogeeks] OS question

2011-08-04 Thread ankit sambyal
What happens when a thread calls exec ?? What happens to the other threads
of the same process ??

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Re: [algogeeks] OS question

2011-08-04 Thread Dipankar Patro
I think the answer would be the thread calling execute will have to wait for
the executed command to exit and then it will proceed.
As for other threads, they shouldn't be affected.

Please do correct me if it is wrong.

On 4 August 2011 20:27, ankit sambyal ankitsamb...@gmail.com wrote:

 What happens when a thread calls exec ?? What happens to the other threads
 of the same process ??

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Re: [algogeeks] OS question

2011-08-04 Thread ankit sambyal
@Dipankar: But all the threads of a process share code and data section. So,
how is it possible that they are not affected ???

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Re: [algogeeks] OS question

2011-08-04 Thread Poised~
Good point.
Let me search a bit on Threads. Will get back asap.

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Re: [algogeeks] OS question

2011-08-04 Thread Azhar Hussain
To elaborate more. New process image will not have the existing threads and
user defined data declared in current process will be wiped out. Parent can
do is to wait for the child status by calling wait().
for example
main()
{
 pid = fork();
if (child) {
exec(ls);   /// here 'ls' will replace the child process
no matter how many threads it has
   }
   else // parent
wait for child

}

-
Azhar.

On Thu, Aug 4, 2011 at 9:50 PM, Azhar Hussain azhar...@gmail.com wrote:

 The *exec* family of functions shall replace the current process image
 with a new process image. It does not matter how many threads you have
 whole process gets replaced with new one.


 -
 Azhar.


 On Thu, Aug 4, 2011 at 8:27 PM, ankit sambyal ankitsamb...@gmail.comwrote:

 What happens when a thread calls exec ?? What happens to the other threads
 of the same process ??

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Re: [algogeeks] OS question

2011-08-04 Thread Azhar Hussain
The *exec* family of functions shall replace the current process image with
a new process image. It does not matter how many threads you have whole
process gets replaced with new one.


-
Azhar.

On Thu, Aug 4, 2011 at 8:27 PM, ankit sambyal ankitsamb...@gmail.comwrote:

 What happens when a thread calls exec ?? What happens to the other threads
 of the same process ??

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Re: [algogeeks] OS question

2011-08-04 Thread ankit sambyal
Thnks Azhar :)
got the point

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[algogeeks] OS book

2011-07-08 Thread zero_cool
Hey coders if anybody has Operating Systems, by William
Stallings please mail me as early as you can.
my e-mail:brajkishoresa...@gmail.com

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[algogeeks] OS

2011-06-27 Thread Nishant Mittal
plz recommend me some good sites for OS interview questions...
Thanx in advance

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Re: [algogeeks] OS

2011-06-27 Thread Navneet Gupta
One such resource
http://placementsindia.blogspot.com/search/label/Operating%20Systems

On Mon, Jun 27, 2011 at 2:01 PM, Nishant Mittal
mittal.nishan...@gmail.com wrote:
 plz recommend me some good sites for OS interview questions...
 Thanx in advance

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Re: [algogeeks] os

2011-06-23 Thread Abhishek Sharma
@rahul: buddy, u can ignore the mail if u don't want to answer (no offense).
Lets not discourage someone from asking questions...

On Tue, Jun 21, 2011 at 11:23 PM, rahul rahulr...@gmail.com wrote:

 If u want us to solve the GATE paper, please attach the paper, we will post
 the solution.

 regards.


 On Tue, Jun 21, 2011 at 11:21 PM, Akshata Sharma 
 akshatasharm...@gmail.com wrote:

 The atomic fetch-and-set x, y instruction unconditionally sets the memory
 location x to 1 and fetches the old value of x n y without allowing any
 intervening access to the memory location x. consider the following
 implementation of P and V functions on a binary semaphore S.

 void P (binary_semaphore *s)
   unsigned y;
   unsigned * = (s—value);
   do
 fetch—and—set x, y;
   while (y)

 void V (binary_semaphore *s)
S—value = 0;
 Which one of the following is true?
 (A) The implementation may not work if context switching is disabled in P
 (B) Instead of using fetch-and —set, a pair of normal load/store can be
 used
 (C) The implementation of V is wrong
 (D) The code does not implement a binary semaphore

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Re: [algogeeks] OS

2011-06-22 Thread Shachindra A C
last year's gate question?

On Tue, Jun 21, 2011 at 11:32 PM, Akshata Sharma
akshatasharm...@gmail.comwrote:

 But, the OS maintains a separate PC (program counter ),stack and A CPU
 register state for a thread . So option A I am not sure is correct, it says
 ONLY..
 scheduling and accounting information is stored for a process right? Can
 you please explain why C is not correct and D is correct?


 On Tue, Jun 21, 2011 at 11:17 PM, rahul rahulr...@gmail.com wrote:

 A, D

 On Tue, Jun 21, 2011 at 11:16 PM, Akshata Sharma 
 akshatasharm...@gmail.com wrote:

  A thread is usually defined as a ‘light weight process’ because an
 operating system (OS) maintains smaller data structures for a thread than
 for a process. In relation to this, which of the followings is TRUE?
 (A) On per-thread basis, the OS maintains only CPU register state
 (B) The OS does not maintain a separate stack for each thread
 (C) On per-thread basis, the OS does not maintain virtual memory state
 (D) On per thread basis, the OS maintains only scheduling and accounting
 information

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Re: [algogeeks] OS

2011-06-22 Thread sachin sharma
@Rahul

Threads within a process share the same virtual memory space but each has a
separate stack, and possibly thread-local storage. this thread local
storage is register and other private data. They are *lightweight* because a
context switch is simply a case of switching the stack pointer and program
counter and restoring other registers, wheras a *process*context switch
involves switching the MMU context as well.

Moreover, communication between threads within a process is
*lightweight* because
they share an address space.

Now

Option A is not right as it says Only register.

Option B is wrong as it directly opposes

Option C is correct as Threads share address space of Process. Virtually
memory is concerned with processes not with Threads

Option D is wrong as it says only scheduling and accounting.


Best Wishes
Sachin Sharma
University of Delhi

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[algogeeks] OS

2011-06-21 Thread Akshata Sharma
 A thread is usually defined as a ‘light weight process’ because an
operating system (OS) maintains smaller data structures for a thread than
for a process. In relation to this, which of the followings is TRUE?
(A) On per-thread basis, the OS maintains only CPU register state
(B) The OS does not maintain a separate stack for each thread
(C) On per-thread basis, the OS does not maintain virtual memory state
(D) On per thread basis, the OS maintains only scheduling and accounting
information

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Re: [algogeeks] OS

2011-06-21 Thread rahul
A, D

On Tue, Jun 21, 2011 at 11:16 PM, Akshata Sharma
akshatasharm...@gmail.comwrote:

  A thread is usually defined as a ‘light weight process’ because an
 operating system (OS) maintains smaller data structures for a thread than
 for a process. In relation to this, which of the followings is TRUE?
 (A) On per-thread basis, the OS maintains only CPU register state
 (B) The OS does not maintain a separate stack for each thread
 (C) On per-thread basis, the OS does not maintain virtual memory state
 (D) On per thread basis, the OS maintains only scheduling and accounting
 information

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[algogeeks] os

2011-06-21 Thread Akshata Sharma
The atomic fetch-and-set x, y instruction unconditionally sets the memory
location x to 1 and fetches the old value of x n y without allowing any
intervening access to the memory location x. consider the following
implementation of P and V functions on a binary semaphore S.

void P (binary_semaphore *s)
  unsigned y;
  unsigned * = (s—value);
  do
fetch—and—set x, y;
  while (y)

void V (binary_semaphore *s)
   S—value = 0;
Which one of the following is true?
(A) The implementation may not work if context switching is disabled in P
(B) Instead of using fetch-and —set, a pair of normal load/store can be used
(C) The implementation of V is wrong
(D) The code does not implement a binary semaphore

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Re: [algogeeks] os

2011-06-21 Thread rahul
If u want us to solve the GATE paper, please attach the paper, we will post
the solution.

regards.

On Tue, Jun 21, 2011 at 11:21 PM, Akshata Sharma
akshatasharm...@gmail.comwrote:

 The atomic fetch-and-set x, y instruction unconditionally sets the memory
 location x to 1 and fetches the old value of x n y without allowing any
 intervening access to the memory location x. consider the following
 implementation of P and V functions on a binary semaphore S.

 void P (binary_semaphore *s)
   unsigned y;
   unsigned * = (s—value);
   do
 fetch—and—set x, y;
   while (y)

 void V (binary_semaphore *s)
S—value = 0;
 Which one of the following is true?
 (A) The implementation may not work if context switching is disabled in P
 (B) Instead of using fetch-and —set, a pair of normal load/store can be
 used
 (C) The implementation of V is wrong
 (D) The code does not implement a binary semaphore

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Re: [algogeeks] OS

2011-06-21 Thread Akshata Sharma
But, the OS maintains a separate PC (program counter ),stack and A CPU
register state for a thread . So option A I am not sure is correct, it says
ONLY..
scheduling and accounting information is stored for a process right? Can you
please explain why C is not correct and D is correct?

On Tue, Jun 21, 2011 at 11:17 PM, rahul rahulr...@gmail.com wrote:

 A, D

 On Tue, Jun 21, 2011 at 11:16 PM, Akshata Sharma 
 akshatasharm...@gmail.com wrote:

  A thread is usually defined as a ‘light weight process’ because an
 operating system (OS) maintains smaller data structures for a thread than
 for a process. In relation to this, which of the followings is TRUE?
 (A) On per-thread basis, the OS maintains only CPU register state
 (B) The OS does not maintain a separate stack for each thread
 (C) On per-thread basis, the OS does not maintain virtual memory state
 (D) On per thread basis, the OS maintains only scheduling and accounting
 information

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[algogeeks] OS

2011-06-19 Thread Akshata Sharma
Two processes, P1 and P2, need to access a critical section of code.
Consider the following synchronization construct used by the processes:

/* P1 */
while (true) {
  wants1 = true;
  while (wants2==true);
/* Critical Section */
  wants1=false;
}
/* Remainder section */

/* P2 */
while (true) {
wants2 = true;
while (wants1==true);
/* Critical Section */
Wants2=false;
}
/* Remainder section */

Here, wants1 and wants2 are shared variables, which are initialized to
false. Which one of the following statements is TRUE about the above
construct?
(A) It does not ensure mutual exclusion.
(B) It does not ensure bounded waiting.
(C) It requires that processes enter the critical section in strict
alternation.
(D) It does not prevent deadlocks, but ensures mutual exclusion.

I think B,C are true. It also prevents deadlock so D is also true, not sure
though. Am I right?

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Re: [algogeeks] OS

2011-06-19 Thread Nishant Mittal
It does *not* prevent deadlock so i think (D) is definitely true.

On Sun, Jun 19, 2011 at 5:15 PM, Akshata Sharma
akshatasharm...@gmail.comwrote:

 Two processes, P1 and P2, need to access a critical section of code.
 Consider the following synchronization construct used by the processes:

 /* P1 */
 while (true) {
   wants1 = true;
   while (wants2==true);
 /* Critical Section */
   wants1=false;
 }
 /* Remainder section */

 /* P2 */
 while (true) {
 wants2 = true;
 while (wants1==true);
 /* Critical Section */
 Wants2=false;
 }
 /* Remainder section */

 Here, wants1 and wants2 are shared variables, which are initialized to
 false. Which one of the following statements is TRUE about the above
 construct?
 (A) It does not ensure mutual exclusion.
 (B) It does not ensure bounded waiting.
 (C) It requires that processes enter the critical section in strict
 alternation.
 (D) It does not prevent deadlocks, but ensures mutual exclusion.

 I think B,C are true. It also prevents deadlock so D is also true, not sure
 though. Am I right?

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Re: [algogeeks] OS

2011-06-19 Thread sanjay ahuja
B and D are true


On Sun, Jun 19, 2011 at 5:43 PM, Nishant Mittal
mittal.nishan...@gmail.com wrote:
 It does not prevent deadlock so i think (D) is definitely true.

 On Sun, Jun 19, 2011 at 5:15 PM, Akshata Sharma akshatasharm...@gmail.com
 wrote:

 Two processes, P1 and P2, need to access a critical section of code.
 Consider the following synchronization construct used by the processes:

 /* P1 */
 while (true) {
   wants1 = true;
   while (wants2==true);
     /* Critical Section */
   wants1=false;
 }
 /* Remainder section */

 /* P2 */
 while (true) {
 wants2 = true;
 while (wants1==true);
 /* Critical Section */
 Wants2=false;
 }
 /* Remainder section */

 Here, wants1 and wants2 are shared variables, which are initialized to
 false. Which one of the following statements is TRUE about the above
 construct?
 (A) It does not ensure mutual exclusion.
 (B) It does not ensure bounded waiting.
 (C) It requires that processes enter the critical section in strict
 alternation.
 (D) It does not prevent deadlocks, but ensures mutual exclusion.

 I think B,C are true. It also prevents deadlock so D is also true, not
 sure though. Am I right?

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-- 
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Analyst, Financing Prime Brokerage
Nomura Securities India Pvt. Ltd

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Re: [algogeeks] OS

2011-06-19 Thread Akshata Sharma
Why is C not true?

On Sun, Jun 19, 2011 at 6:31 PM, sanjay ahuja sanjayahuja.i...@gmail.comwrote:

 B and D are true


 On Sun, Jun 19, 2011 at 5:43 PM, Nishant Mittal
 mittal.nishan...@gmail.com wrote:
  It does not prevent deadlock so i think (D) is definitely true.
 
  On Sun, Jun 19, 2011 at 5:15 PM, Akshata Sharma 
 akshatasharm...@gmail.com
  wrote:
 
  Two processes, P1 and P2, need to access a critical section of code.
  Consider the following synchronization construct used by the processes:
 
  /* P1 */
  while (true) {
wants1 = true;
while (wants2==true);
  /* Critical Section */
wants1=false;
  }
  /* Remainder section */
 
  /* P2 */
  while (true) {
  wants2 = true;
  while (wants1==true);
  /* Critical Section */
  Wants2=false;
  }
  /* Remainder section */
 
  Here, wants1 and wants2 are shared variables, which are initialized to
  false. Which one of the following statements is TRUE about the above
  construct?
  (A) It does not ensure mutual exclusion.
  (B) It does not ensure bounded waiting.
  (C) It requires that processes enter the critical section in strict
  alternation.
  (D) It does not prevent deadlocks, but ensures mutual exclusion.
 
  I think B,C are true. It also prevents deadlock so D is also true, not
  sure though. Am I right?
 
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Re: [algogeeks] OS

2011-06-19 Thread sanjay ahuja
where does it ensure that if P1 has first executed critical section
then it will get chance to execute critical section only after P2 has
executed critical section once.

If it is strict alternation then it is ensuring bounded waiting!


On Sun, Jun 19, 2011 at 7:01 PM, Akshata Sharma
akshatasharm...@gmail.com wrote:
 Why is C not true?

 On Sun, Jun 19, 2011 at 6:31 PM, sanjay ahuja sanjayahuja.i...@gmail.com
 wrote:

 B and D are true


 On Sun, Jun 19, 2011 at 5:43 PM, Nishant Mittal
 mittal.nishan...@gmail.com wrote:
  It does not prevent deadlock so i think (D) is definitely true.
 
  On Sun, Jun 19, 2011 at 5:15 PM, Akshata Sharma
  akshatasharm...@gmail.com
  wrote:
 
  Two processes, P1 and P2, need to access a critical section of code.
  Consider the following synchronization construct used by the processes:
 
  /* P1 */
  while (true) {
    wants1 = true;
    while (wants2==true);
      /* Critical Section */
    wants1=false;
  }
  /* Remainder section */
 
  /* P2 */
  while (true) {
  wants2 = true;
  while (wants1==true);
  /* Critical Section */
  Wants2=false;
  }
  /* Remainder section */
 
  Here, wants1 and wants2 are shared variables, which are initialized to
  false. Which one of the following statements is TRUE about the above
  construct?
  (A) It does not ensure mutual exclusion.
  (B) It does not ensure bounded waiting.
  (C) It requires that processes enter the critical section in strict
  alternation.
  (D) It does not prevent deadlocks, but ensures mutual exclusion.
 
  I think B,C are true. It also prevents deadlock so D is also true, not
  sure though. Am I right?
 
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Re: [algogeeks] OS

2011-06-19 Thread Wladimir Tavares
One of two process can be in starvation!

Wladimir Araujo Tavares
*Federal University of Ceará

*




On Sun, Jun 19, 2011 at 10:49 AM, sanjay ahuja
sanjayahuja.i...@gmail.comwrote:

 where does it ensure that if P1 has first executed critical section
 then it will get chance to execute critical section only after P2 has
 executed critical section once.

 If it is strict alternation then it is ensuring bounded waiting!


 On Sun, Jun 19, 2011 at 7:01 PM, Akshata Sharma
 akshatasharm...@gmail.com wrote:
  Why is C not true?
 
  On Sun, Jun 19, 2011 at 6:31 PM, sanjay ahuja 
 sanjayahuja.i...@gmail.com
  wrote:
 
  B and D are true
 
 
  On Sun, Jun 19, 2011 at 5:43 PM, Nishant Mittal
  mittal.nishan...@gmail.com wrote:
   It does not prevent deadlock so i think (D) is definitely true.
  
   On Sun, Jun 19, 2011 at 5:15 PM, Akshata Sharma
   akshatasharm...@gmail.com
   wrote:
  
   Two processes, P1 and P2, need to access a critical section of code.
   Consider the following synchronization construct used by the
 processes:
  
   /* P1 */
   while (true) {
 wants1 = true;
 while (wants2==true);
   /* Critical Section */
 wants1=false;
   }
   /* Remainder section */
  
   /* P2 */
   while (true) {
   wants2 = true;
   while (wants1==true);
   /* Critical Section */
   Wants2=false;
   }
   /* Remainder section */
  
   Here, wants1 and wants2 are shared variables, which are initialized
 to
   false. Which one of the following statements is TRUE about the above
   construct?
   (A) It does not ensure mutual exclusion.
   (B) It does not ensure bounded waiting.
   (C) It requires that processes enter the critical section in strict
   alternation.
   (D) It does not prevent deadlocks, but ensures mutual exclusion.
  
   I think B,C are true. It also prevents deadlock so D is also true,
 not
   sure though. Am I right?
  
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Re: [algogeeks] OS

2011-06-19 Thread Wladimir Tavares
If the operation (want = false) is not atomic, we can not mutual
exclusion.Certo?

Wladimir Araujo Tavares
*Federal University of Ceará

*




On Sun, Jun 19, 2011 at 11:01 AM, Wladimir Tavares wladimir...@gmail.comwrote:

 One of two process can be in starvation!

 Wladimir Araujo Tavares
 *Federal University of Ceará

 *





 On Sun, Jun 19, 2011 at 10:49 AM, sanjay ahuja sanjayahuja.i...@gmail.com
  wrote:

 where does it ensure that if P1 has first executed critical section
 then it will get chance to execute critical section only after P2 has
 executed critical section once.

 If it is strict alternation then it is ensuring bounded waiting!


 On Sun, Jun 19, 2011 at 7:01 PM, Akshata Sharma
 akshatasharm...@gmail.com wrote:
  Why is C not true?
 
  On Sun, Jun 19, 2011 at 6:31 PM, sanjay ahuja 
 sanjayahuja.i...@gmail.com
  wrote:
 
  B and D are true
 
 
  On Sun, Jun 19, 2011 at 5:43 PM, Nishant Mittal
  mittal.nishan...@gmail.com wrote:
   It does not prevent deadlock so i think (D) is definitely true.
  
   On Sun, Jun 19, 2011 at 5:15 PM, Akshata Sharma
   akshatasharm...@gmail.com
   wrote:
  
   Two processes, P1 and P2, need to access a critical section of code.
   Consider the following synchronization construct used by the
 processes:
  
   /* P1 */
   while (true) {
 wants1 = true;
 while (wants2==true);
   /* Critical Section */
 wants1=false;
   }
   /* Remainder section */
  
   /* P2 */
   while (true) {
   wants2 = true;
   while (wants1==true);
   /* Critical Section */
   Wants2=false;
   }
   /* Remainder section */
  
   Here, wants1 and wants2 are shared variables, which are initialized
 to
   false. Which one of the following statements is TRUE about the above
   construct?
   (A) It does not ensure mutual exclusion.
   (B) It does not ensure bounded waiting.
   (C) It requires that processes enter the critical section in strict
   alternation.
   (D) It does not prevent deadlocks, but ensures mutual exclusion.
  
   I think B,C are true. It also prevents deadlock so D is also true,
 not
   sure though. Am I right?
  
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Re: [algogeeks] OS galvin sol..

2011-01-21 Thread jayapriya surendran
wow..thank you so much

On Wed, Jan 19, 2011 at 2:08 PM, LALIT SHARMA lks.ru...@gmail.com wrote:



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Re: [algogeeks] OS galvin sol..

2011-01-21 Thread Sreeprasad Govindankutty
Thanks so much

On Wed, Jan 19, 2011 at 4:20 AM, jayapriya surendran priya7...@gmail.comwrote:

 wow..thank you so much


 On Wed, Jan 19, 2011 at 2:08 PM, LALIT SHARMA lks.ru...@gmail.com wrote:



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Re: [algogeeks] OS galvin sol..

2011-01-21 Thread Anand
It's really good. Thanks a lot

On Fri, Jan 21, 2011 at 8:24 AM, Sreeprasad Govindankutty 
sreeprasad...@gmail.com wrote:

 Thanks so much


 On Wed, Jan 19, 2011 at 4:20 AM, jayapriya surendran 
 priya7...@gmail.comwrote:

 wow..thank you so much


 On Wed, Jan 19, 2011 at 2:08 PM, LALIT SHARMA lks.ru...@gmail.comwrote:



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 IIIT Allahabad (Amethi Capmus),
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Re: [algogeeks] OS galvin sol..

2011-01-21 Thread LALIT SHARMA
My Pleasure !!
On Fri, Jan 21, 2011 at 11:22 PM, Anand anandut2...@gmail.com wrote:

 It's really good. Thanks a lot


 On Fri, Jan 21, 2011 at 8:24 AM, Sreeprasad Govindankutty 
 sreeprasad...@gmail.com wrote:

 Thanks so much


 On Wed, Jan 19, 2011 at 4:20 AM, jayapriya surendran priya7...@gmail.com
  wrote:

 wow..thank you so much


 On Wed, Jan 19, 2011 at 2:08 PM, LALIT SHARMA lks.ru...@gmail.comwrote:



 --
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 IIIT Allahabad (Amethi Capmus),
 6th Sem.

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Re: [algogeeks] OS galvin sol..

2011-01-21 Thread rahul rai
can u give me sipser solution mannual?


On 1/21/11, Sreeprasad Govindankutty sreeprasad...@gmail.com wrote:
 Thanks so much

 On Wed, Jan 19, 2011 at 4:20 AM, jayapriya surendran
 priya7...@gmail.comwrote:

 wow..thank you so much


 On Wed, Jan 19, 2011 at 2:08 PM, LALIT SHARMA lks.ru...@gmail.com wrote:



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Re: [algogeeks] os problem

2010-07-22 Thread topojoy biswas
But you dont need a swap filesystem right?

On Thu, Jul 22, 2010 at 8:41 AM, Anand anandut2...@gmail.com wrote:

 Yes you do need virtual memory even if you have 4GB of RAM. Because if you
 do not have virtual memory, you could not have uniform addressing. and that
 prevents you creating the final elf file for each process. B'cos while
 compiling the program you don;t know the actual physical address your
 program is going to reside during execution.



 On Tue, Jul 20, 2010 at 11:46 AM, divya sweetdivya@gmail.com wrote:

 You have 4GB ram, and at any time you have only 2 processes of 10mb
 each. so do you need any virtual memory for it?

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[algogeeks] os problem

2010-07-21 Thread divya
You have 4GB ram, and at any time you have only 2 processes of 10mb
each. so do you need any virtual memory for it?

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Re: [algogeeks] os problem

2010-07-21 Thread Anand
Yes you do need virtual memory even if you have 4GB of RAM. Because if you
do not have virtual memory, you could not have uniform addressing. and that
prevents you creating the final elf file for each process. B'cos while
compiling the program you don;t know the actual physical address your
program is going to reside during execution.



On Tue, Jul 20, 2010 at 11:46 AM, divya sweetdivya@gmail.com wrote:

 You have 4GB ram, and at any time you have only 2 processes of 10mb
 each. so do you need any virtual memory for it?

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Re: [algogeeks] OS problems

2010-07-01 Thread Anand
if there are 32 such frames of 8 X 1024 then the logical address will be
(10+5)15 as pointed out by Harit.

On Thu, Jul 1, 2010 at 8:57 AM, sharad kumar sharad20073...@gmail.comwrote:

 i think harit's answer is correct regarding ques 2 plzz someone comment on
 this

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Re: [algogeeks] OS problems

2010-06-25 Thread harit agarwal
1. the virtual memory size depends on the page size that the system is
using...
2. logical address=5+10=15 bits + (some modifying bits if they are present
like modified,copied etc..)

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Re: [algogeeks] OS problems

2010-06-19 Thread Amit Jaspal
@ above

I think it is because of the heap size . The Heap corresponding to dynamic
memory allocation grows and merges with the stack section of the process.

Correct me if I am wrong.
And if was only because of calloc() , then will malloc work?

Can we allocate 1gb dynamically using malloc()??


On Sat, Jun 19, 2010 at 10:14 AM, harit agarwal agarwalha...@gmail.comwrote:

 @amit

 1. calloc gives contiguos allocated space and it is not necessary that  it
 can find 1gb in a row that's why it failed after allocating some memory...
  it is not necessary that it will always allocate 800mb of space as in this
 case...


 2. whenever a process is executed in critical sectionit is means that
 it raises it execution level so that it can't be interrupted while the other
 processors are still on the same execution level they can be interrupted

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Re: [algogeeks] OS problems

2010-06-19 Thread sharad kumar
for 1
other reasons apart 4m d 1 told by harit are
1)in every os,a user has maximum space allocated to him according to his
previlege so ... may be it is exceeding that maximum capacity
2)it may be possible that it has exceeded total space available to whole os
i.e it may be smaller system having much less physical memeory

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Re: [algogeeks] OS problems

2010-06-19 Thread harit agarwal
yes you can allocate 1gb using malloc but it also depends on how much heap
size is available to you..
if you try 2gb then more chances are it won't allocate because of heap is
exhausted..

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Re: [algogeeks] OS problems

2010-06-19 Thread Anand
In SMP operating sytem provides spinlock to execute critical section of code
that is shared among various processors. Spinlocks keeps every other
processors just to spin around and there by prevents them from generating
interrupts which could interrupt the processor which is executing the
critical section.
On Thu, Jun 17, 2010 at 12:39 PM, amit amitjaspal...@gmail.com wrote:

 1. a mad user tries to allocate 1 gb memory using calloc.
 but the program fails after allocationg about 800mg(appx. i dont
 remember). Tell me what could have gone wrong?

 2.
 We know disabling interrupts works only if it is single processor(i.e
 local disabling of interrupts).

 Consider this case where we have a SMP(symmetric multi proc) the
 processor. Processor-1 wants to perform some critical operation so it
 disables all the interrupts.

 What will happen when processor-2 throws an interrupt.

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Re: [algogeeks] OS problems

2010-06-19 Thread harit agarwal
@amit

i think your query is answered by varun..as each process do system call to
allocate memory so it is exhausting the memory for all the processesas
all  processes are having the same interface...

@sharad

1.i don't think priviliges affect the user address spaceit tells that in
which way the user can use that space
2.as you said may be whole memory available to OS 1 gb-this is the
extreme case and also i don't think such kind of systems exist now...

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Re: [algogeeks] OS problems

2010-06-18 Thread jaladhi dave
can you explain what you meant when you said the program fails after
allocationg about 800mg(appx. i dont remember).
This is the excerpt from calloc man page, Calloc will either fail or succeed
but there is no way you can tell so much was alloted and then it failed.
*Return Value***For calloc() and malloc(), the value returned is a pointer
to the allocated memory, which is suitably aligned for any kind of variable,
or NULL if the request fails.

On Fri, Jun 18, 2010 at 1:09 AM, amit amitjaspal...@gmail.com wrote:

 1. a mad user tries to allocate 1 gb memory using calloc.
 but the program fails after allocationg about 800mg(appx. i dont
 remember). Tell me what could have gone wrong?

 2.
 We know disabling interrupts works only if it is single processor(i.e
 local disabling of interrupts).

 Consider this case where we have a SMP(symmetric multi proc) the
 processor. Processor-1 wants to perform some critical operation so it
 disables all the interrupts.

 What will happen when processor-2 throws an interrupt.

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Re: [algogeeks] OS problems

2010-06-18 Thread Amit Jaspal
It means the program crashed while it was trying to allocate more memory .
Now can u guess why that happened?

On Fri, Jun 18, 2010 at 1:29 PM, jaladhi dave jaladhi.k.d...@gmail.comwrote:

 can you explain what you meant when you said the program fails after
 allocationg about 800mg(appx. i dont remember).
 This is the excerpt from calloc man page, Calloc will either fail or
 succeed but there is no way you can tell so much was alloted and then it
 failed.
 *Return Value***For calloc() and malloc(), the value returned is a pointer
 to the allocated memory, which is suitably aligned for any kind of variable,
 or NULL if the request fails.

 On Fri, Jun 18, 2010 at 1:09 AM, amit amitjaspal...@gmail.com wrote:

 1. a mad user tries to allocate 1 gb memory using calloc.
 but the program fails after allocationg about 800mg(appx. i dont
 remember). Tell me what could have gone wrong?

 2.
 We know disabling interrupts works only if it is single processor(i.e
 local disabling of interrupts).

 Consider this case where we have a SMP(symmetric multi proc) the
 processor. Processor-1 wants to perform some critical operation so it
 disables all the interrupts.

 What will happen when processor-2 throws an interrupt.

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Re: [algogeeks] OS problems

2010-06-18 Thread harit agarwal
@amit

1. calloc gives contiguos allocated space and it is not necessary that  it
can find 1gb in a row that's why it failed after allocating some memory...
 it is not necessary that it will always allocate 800mb of space as in this
case...


2. whenever a process is executed in critical sectionit is means that it
raises it execution level so that it can't be interrupted while the other
processors are still on the same execution level they can be interrupted

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[algogeeks] OS problems

2010-06-17 Thread amit
1. a mad user tries to allocate 1 gb memory using calloc.
but the program fails after allocationg about 800mg(appx. i dont
remember). Tell me what could have gone wrong?

2.
We know disabling interrupts works only if it is single processor(i.e
local disabling of interrupts).

Consider this case where we have a SMP(symmetric multi proc) the
processor. Processor-1 wants to perform some critical operation so it
disables all the interrupts.

What will happen when processor-2 throws an interrupt.

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Re: [algogeeks] OS doubt

2010-06-15 Thread Anand
Uninitialized global variables are stored in .bss section of the process
memory and initialised global variables are stored in .data section of the
memory. In the linking stage, they get the actually physical address. But
since x and y are local variables they are just stored in stack while
execution and will get flushed out later from stack after its execution. So
they don't have any physical address for debugging.

On Sat, Jun 12, 2010 at 1:22 AM, amit amitjaspal...@gmail.com wrote:

 OS doubt:

 I have read many times that say a 24 KB process enters the Main Memory
 selected by the Long Term Scheduler.
 But I don't understand what it exactly means.
 As far as I know Process consists of ( Code + Data(Static) +
 Stack(Local Data) + Heap)

 So doubt1: Is this 24 KB the size of this whole process or just the
 size of the code segment.

 doubt2: Now lets say this process starts getting executed by the
 CPU ,Suppose the main() contains
main(){
int x;
int y;
x=10;
...
}
So x,y will be allocated the memory in the Stack.
But when x=10 is encountered , how will the CPU know the
 address of
 x. In short how is x accessed??


 doubt 3: If x and y are just address of a memory location in the
 stack , can their logical address be determined by the compiler or it
 will be generated by the CPU??

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[algogeeks] OS doubt

2010-06-12 Thread amit
OS doubt:

I have read many times that say a 24 KB process enters the Main Memory
selected by the Long Term Scheduler.
But I don't understand what it exactly means.
As far as I know Process consists of ( Code + Data(Static) +
Stack(Local Data) + Heap)

So doubt1: Is this 24 KB the size of this whole process or just the
size of the code segment.

doubt2: Now lets say this process starts getting executed by the
CPU ,Suppose the main() contains
main(){
int x;
int y;
x=10;
...
}
So x,y will be allocated the memory in the Stack.
But when x=10 is encountered , how will the CPU know the 
address of
x. In short how is x accessed??


doubt 3: If x and y are just address of a memory location in the
stack , can their logical address be determined by the compiler or it
will be generated by the CPU??

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Re: [algogeeks] os

2010-06-10 Thread sharad kumar
@sharad :

Mutex can be released only by thread that had acquired it, while you can
signal semaphore from any other thread (or process), so semaphores are more
suitable for some synchronization problems like producer-consumer.

One Windows binary semaphores are more like event objects then mutants.

On Thu, Jun 10, 2010 at 9:19 PM, sharad sharad20073...@gmail.com wrote:

 What is the difference between Mutex and Binary semaphore

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Re: [algogeeks] os

2010-06-10 Thread sharad kumar
@sharad but when it is binary semaphore then only one process is accessing
the resource,rest all are blockedwhich means that only that process who
locked bin. sem will unlock it .plzzz correct me if i m wrong

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