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Pointer addresses in multiples of 32 ?

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# 1  
Old 06-01-2010
Pointer addresses in multiples of 32 ?
1. Why are the pointers' addresses every 32 ?
2. Am I correct in stating that memset is writing to memory that is not allocated to any of the 3 pointers ? Is it writing to memory in between the pointers ?
3. Are the 3 pointers contiguous in memory ?
4. I only allocated 10 bytes for each pointer. So how can "a" be 73 chars long etc. ?

Code:
size_t bufsize = 10;

char *a = malloc(bufsize);
memset(a, 'a', bufsize - 1);
a[bufsize] = '\0';

char *b = malloc(bufsize);
memset(b, 'b', bufsize - 1);
b[bufsize] = '\0';

char *c = malloc(bufsize);
memset(c, 'c', bufsize - 1);
c[bufsize] = '\0';

size_t d = c - a;

memset(a, '-', d);
printf("a = %s => address: %i => number of chars: %i\n", a, a, strlen(a));
printf("b = %s => address: %i => number of chars: %i\n", b, b, strlen(b));
printf("c = %s => address: %i => number of chars: %i\n", c, c, strlen(c));
printf("d = %i\n", d);

Quote:
a = ----------------------------------------------------------------ccccccccc => address: 6299616 => number of chars: 73
b = --------------------------------ccccccccc => address: 6299648 => number of chars: 41
c = ccccccccc => address: 6299680 => number of chars: 9
d = 64
# 2  
Old 06-01-2010
Quote:
Originally Posted by cyler
Code:
size_t d = c - a;

memset(a, '-', d);

you can't mess about with memory like this and get away with it.
it may work once or twice, but it will cause a seg fault one day.

you cannot make assumptions about how malloc works
or where memory is allocated.
:-)
# 3  
Old 06-01-2010
Quote:
Originally Posted by cyler
1. Why are the pointers' addresses every 32 ?
That depends on the bitness of your machine...if it is 32-bit then pointers will store 4 bytes and if it is 64-bit then pointers will store 8 bytes.
Quote:
Originally Posted by cyler
2. Am I correct in stating that memset is writing to memory that is not allocated to any of the 3 pointers ? Is it writing to memory in between the pointers ?
Depends on what memory locations a b and c are pointing to...just dont assume anything like that.
Quote:
Originally Posted by cyler
3. Are the 3 pointers contiguous in memory ?
Again dont assume anything...be satisfied with what the os gives you.
Quote:
Originally Posted by cyler
4. I only allocated 10 bytes for each pointer. So how can "a" be 73 chars long etc. ?
See the line in red below for the answer.
Code:
size_t bufsize = 10;

char *a = malloc(bufsize);
memset(a, 'a', bufsize - 1);
a[bufsize] = '\0';

char *b = malloc(bufsize);
memset(b, 'b', bufsize - 1);
b[bufsize] = '\0';

char *c = malloc(bufsize);
memset(c, 'c', bufsize - 1);
c[bufsize] = '\0';

size_t d = c - a;

memset(a, '-', d);
printf("a = %s => address: %i => number of chars: %i\n", a, a, strlen(a));
printf("b = %s => address: %i => number of chars: %i\n", b, b, strlen(b));
printf("c = %s => address: %i => number of chars: %i\n", c, c, strlen(c));
printf("d = %i\n", d);

# 4  
Old 06-01-2010
Quote:
Originally Posted by shamrock
Depends on what memory locations a b and c are pointing to...just dont assume anything like that.
I am trying to follow this article: http://www.w00w00.org/files/articles/heaptut.txt

This article's example successfully uses memset to overwrite a pointer's address while making it overflow to overwrite another address. (Just wondering if both pointers must be contiguous in memory for this to happen.)

So basically:
1. Depends on 32 vs 64 bit machines.
2. No, it writes to unused memory in the memory allocated for each pointer.
3. ???
4. Has to do with 1 & 2.
# 5  
Old 06-02-2010
Quote:
Originally Posted by cyler
1. Why are the pointers' addresses every 32 ?
2. Am I correct in stating that memset is writing to memory that is not allocated to any of the 3 pointers ? Is it writing to memory in between the pointers ?
3. Are the 3 pointers contiguous in memory ?
4. I only allocated 10 bytes for each pointer. So how can "a" be 73 chars long etc. ?

Code:
size_t bufsize = 10;

char *a = malloc(bufsize);
memset(a, 'a', bufsize - 1);
a[bufsize] = '\0';

char *b = malloc(bufsize);
memset(b, 'b', bufsize - 1);
b[bufsize] = '\0';

char *c = malloc(bufsize);
memset(c, 'c', bufsize - 1);
c[bufsize] = '\0';

size_t d = c - a;

memset(a, '-', d);
printf("a = %s => address: %i => number of chars: %i\n", a, a, strlen(a));
printf("b = %s => address: %i => number of chars: %i\n", b, b, strlen(b));
printf("c = %s => address: %i => number of chars: %i\n", c, c, strlen(c));
printf("d = %i\n", d);

The following are also wrong:

Code:
char *a = malloc(bufsize);
memset(a, 'a', bufsize - 1);
a[bufsize] = '\0';  <--- memory corruption here!

The code "a[10]" refers to the 11th element of the array since the index starts at zero.
# 6  
Old 06-02-2010
On my BSD machine I get all sorts of differences...

Code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, char ** argv){
        
    if(argc == 1) return 0;
    size_t sz = atoi(argv[1]);

    char *ab = malloc(sz);
    char *ap = malloc(sz);

    printf("size =  %d\n", (int)sz);
    printf("ab =  %lu\n", (long) ab );
    printf("ap =  %lu\n", (long) ap );
    printf("diff =  %td\n", ap - ab);
    return 0;

}

Code:
$ ./memdiff 1  
size =  1
ab =  34369184004
ap =  34369184006
diff =  2
$ ./memdiff 2
size =  2
ab =  34369184004
ap =  34369184006
diff =  2
$ ./memdiff 10
size =  10
ab =  34369183808
ap =  34369183824
diff =  16
$ ./memdiff 32
size =  32
ab =  34369183808
ap =  34369183840
diff =  32

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