bit to all the remaining positions
making it 0x (-1). You can confirm it while setting bit.bit1 to 0
-Dinesh Bansal
On Tue, Aug 9, 2011 at 12:24 PM, Rohit Srivastava access2ro...@gmail.comwrote:
#includestdio.h
#includeconio.h
int main()
{
struct value
{
int bit1:1
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);
NewWord = 0;
}
}
ii++;
}
/* Now print the words from stack */
while(1) {
ii = pop();
printf(%s ,(char *)InStr[ii]);
if(ii == 0) break;
}
}
-Dinesh Bansal
On Wed, Jul 6, 2011 at 9:49 PM, Navneet Gupta navneetn...@gmail.com wrote
;
}
}
}
for(i=0; i size;i++)
printf(%c,input[i]);
}
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in. Since
medium priority tasks does not need R, they can interrupt L. Now if H
also need the same resource R, it has to wait until L is done with it.
So H gets blocked. This is called priority inversion.
-Dinesh Bansal
On Mon, Jun 20, 2011 at 10:35 PM, ricky moon.afr...@gmail.com wrote:
In priority
Thanks Guys I got it.
@balaji... you are right.. it will work just fine.
-Dinesh Bansal
On Fri, Jun 10, 2011 at 10:22 PM, Vetri Balaji vetribal...@gmail.com wrote:
int flip(int j,int k,int n)
{
int t1=(1j)-1;
int t2=(1k)-1;
t1=t2^t1;
return n^t1;
}
correct me if im wrong
On Fri
How do you reverse the bits between j to k in a 32 bit integer.
For e.g.:
n = 11100011; j = 2 and k = 5
output: 1101 (bits from 2 to 5 are reversed.)
n = 11010110; j = 1 and k = 5
output: 11101000
O(1) method is preferred.
Thanks,
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Medium | 5
Low | 2
Calculate total weight = 8*4+5*2+2*1 = 44
Now assign resource based on priority:
Priority = high, Resource = 4 * 100 / 44
Priority = medium, Resource = 2 * 100 / 44
Priority= low, Resource = 1 * 100 / 44
Hope it makes more clear.
-Dinesh Bansal
On Thu
type
On Tue, Feb 15, 2011 at 12:36 PM, rahul rahulr...@gmail.com wrote:
char *start is const char *start, pointer to const char, u can't
derefernce it and change it.
take a char start[256].try this.
On Tue, Feb 15, 2011 at 12:31 PM, dinesh bansal bansal...@gmail.comwrote:
Hi All,
Can you
+=2,j+=2)
{
start[i] = hex_buf[j];
start[i+1] = hex_buf[j+1];
}
}
Thanks,
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The Law of Win says
Hi Guys,
There are some questions asked to me:
1. How do you print the SLL in reverse order. List should not be changed.
2. Two SLLs are merging at one point, how can you find out efficiently.
Thanks
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Hi juver++,
Ya.. recurrsive function call was a good idea.. thanks.
Regrading second question, there are two SLLs L1 and L2, at some node both
lists nodes point to the same node and create single linked list.
hope I am clear.
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The Law of Win says
.
Use External sorting.
Divide the file data into smaller chunks. Sort the chunks of data one at a
time and save them back to file.
Then get top values from the chunks and sort them again till you get top 10
values.
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The Law of Win says, Let's not do it your way or my way; let's do
If we rewrite question in terms of Probability, call to foo2() depends on
two events:
1. (E1) A B, probablity 75%.
2. (E2) C D, again probability 75%.
Probability (E) = Prob(E1) * Prob(E2) = 75/100 * 75/100 * 5000 = 2812.50
times.
Correct me if wrong.
- Dinesh Bansal
On Wed, Dec 15, 2010
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file contents
in main memory. Can somebody help me with algorithm or pseudo code?
Thanks in advance.
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On Mon, Dec 21, 2009 at 6:47 PM, Linus Probert linus.prob...@gmail.comwrote:
If the numbers are unique you could use a bitmap-sort this way you could
easily read just parts of the file at a time.
If they aren't unique it gets a bit trickier.
/L
dinesh bansal wrote:
Hi All,
Suppose I
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at their specific locations. At the end, display
the nodes from the array.
Thanks,
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The Law of Win says, Let's not do it your way or my way; let's do
by adding an element to the set that is
already there (O(n) expected time to build the set).
Ralph Boland
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The Law of Win says, Let's not do it your way or my way; let's do it the
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