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Old 12-11-2012, 05:08 AM   #1
manoj7410
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Lightbulb Getting unpredictible output by memcpy.


hi..
I was trying to copy one address's data to another location by using memcpy, but getting unpredictible output.. code is :

#include<stdio.h>
int main()
{
int *i,j,k = 10;
i = &k;
// *(&j) = *i;
memcpy (&j,i,10);
printf("i = %d, j = %d\n",*i,j);
return 0;
}

value of *i is giving "segmentation fault" and sometimes a positive value or 0.
Don't know how. Please put some light on it.

Thanks
 
Old 12-11-2012, 05:48 AM   #2
millgates
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What is the program supposed to do?

What does following statement do?
Code:
memcpy (&j,i,10);
 
Old 12-11-2012, 06:06 AM   #3
manoj7410
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Quote:
Originally Posted by millgates View Post
What is the program supposed to do?

What does following statement do?
Code:
memcpy (&j,i,10);
"i" is an integer pointer and m trying to copy the data available at i to the address of j.
j is working fine, but *i is giving seg fault or 0 or some negative value.
 
Old 12-11-2012, 06:10 AM   #4
millgates
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Code:
memcpy (&j,i,10);
OK, but why 10?
 
Old 12-11-2012, 06:14 AM   #5
manoj7410
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Unhappy

Quote:
Originally Posted by millgates View Post
Code:
memcpy (&j,i,10);
OK, but why 10?
just 10 bytes, because I dont know the concept behind it.
 
Old 12-11-2012, 06:19 AM   #6
millgates
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But you're trying to copy 10 bytes to an int which size might vary depending on your platform but I would bet my Slackware install disc that it is less than 10 bytes. How is it supposed to fit there?
 
Old 12-11-2012, 06:19 AM   #7
linosaurusroot
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Your integers are probably 4 bytes, not 10.

Code:
memcpy (&j, i, sizeof(int));
 
Old 12-11-2012, 06:24 AM   #8
manoj7410
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Quote:
Originally Posted by linosaurusroot View Post
Your integers are probably 4 bytes, not 10.

Code:
memcpy (&j, i, sizeof(int));
4 bytes I tried and was working fine, but if I give 10 bytes then, why i is being disturbed..?
 
Old 12-11-2012, 06:31 AM   #9
millgates
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Quote:
Originally Posted by manoj7410 View Post
4 bytes I tried and was working fine, but if I give 10 bytes then, why i is being disturbed..?
because i points to k, and (and I think this is unspecified, though) probably k is stored in memory right after j. So, when you memcpy 10 bytes to j, the first 4 bytes (assuming your int is 4 bytes, which it most likely is) will be written to j, the next 4 bytes will rewrite k (which is where i points to, and the last two bytes will try to overwrite whatever happens to be there in the memory, perhaps some instructions or something like that. Of course the system does not like that.
 
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Old 12-11-2012, 06:48 AM   #10
linosaurusroot
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A guide such as "Smashing the stack for fun and profit" explains what happens when you overflow the bounds of your automatic variables and damage other data on the stack. When programming in C you should take great care not to do this. In fact that's one of the major drawbacks of C.
 
Old 12-11-2012, 07:24 AM   #11
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Quote:
Originally Posted by manoj7410 View Post
but if I give 10 bytes then, why i is being disturbed..?
Code:
int main()
{
        int *i,j,k = 10;
Assuming optimization is off, i, j and k are stored on the stack, almost certainly together and in a sequence that is up to the whim of the compiler and is not predictable.

Code:
memcpy (&j,i,10);
You overwrite the 4 bytes of j and you overwrite whatever six bytes follow j. So if i happens to be directly after j, you overwrite all four bytes of i (and two more beyond). If k is after j and i after k, then you overwrite all four bytes of k and two bytes of i.

If you overwrite i with something that doesn't happen to be a valid address, then the next use of *i will seg fault.

Last edited by johnsfine; 12-11-2012 at 07:26 AM.
 
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Old 12-11-2012, 09:24 PM   #12
manoj7410
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I got it now, Thanks to all..
 
  


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