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
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
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
Hi All,
Please suggest any one, group or forums related to Operating System
and Network like algogeeks.
Thanks
Vivek P
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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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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
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
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
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
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
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
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
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,
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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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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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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How many processes are created in this snippet?
Main()
{
Fork();
Fork() fork () || fork ();
Fork ();
}
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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 ();
}
--
You
a. 15
b. 19
c. 21
d. 27
e. 31
these are the only options.
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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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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,
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
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,
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,
20 is not in option ..so whats the answer??
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@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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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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thnks everyone...
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@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:
@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
How do you make sure to unlock a mutex which was locked in a thread that
dies/terminates?
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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?
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Kamakshi
kamakshi...@gmail.com
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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
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
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
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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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
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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Raghavendhra
changing the face can change nothing .. but facing the change can change
everything
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@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.
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Raghavendhra
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 ...
Fastest IPC mechanism is
1. ?shared memory
2. ?pipes
3. ?named pipes
4. ?Semaphores
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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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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
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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
*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.
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. ?
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
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
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
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
What happens when a thread calls exec ?? What happens to the other threads
of the same process ??
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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
@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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Good point.
Let me search a bit on Threads. Will get back asap.
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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); ///
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
Thnks Azhar :)
got the point
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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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plz recommend me some good sites for OS interview questions...
Thanx in advance
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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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@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.
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
@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
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
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
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
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
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
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)
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
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
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
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
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
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
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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IIIT Allahabad (Amethi Capmus),
6th Sem.
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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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6th Sem.
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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
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
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
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
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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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
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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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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@ 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()??
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
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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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
@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
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()
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
@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
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
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
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
@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
@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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