malloc() allocs memory already allocated to something else - help

Hi all,

I'm trying to readin a 2D array of long integers, of unknown row and col size to start with.

Here is the relevant part of the code

Code:

---

...
...
long int **datacols;

colstofindcount=6;
datacols=(long int **)malloc(colstofindcount*sizeof(long int *));
if (datacols==NULL){
printf("Unable to allocate memory to datacols\n");
exit(1);
}
colslen=26416;
for(i=0;i<noofcolstofind;i++){
*(datacols+i)=(long int *)malloc(colslen*sizeof(long int));
if (*(datacols+i)==NULL){
printf("Unable to allocate memory\n");
exit(1);
}
} /* end of for loop allocating memory to all cols*/

---

Here is the memory map of what the compiler allocates


datacols= 0x94f5968

I expect datacols to be allocated 24 bytes as sizeof(long int) on my machine is 4 bytes

So I expect the memory from 0x94f5968 to 0x94f598c to be allocated, as is shown below.

(datacols+0) 0x94f5968
(datacols+1) 0x94f596c
(datacols+2) 0x94f5970
(datacols+3) 0x94f5974
(datacols+4) 0x94f5978
(datacols+5) 0x94f597c

If we see the blocks of memory allocated in the for loop, I see

*(datacols+0) 0x94f5978
*(datacols+1) 0x950f638
*(datacols+2) 0x95292f8
*(datacols+3) 0x9542fb8
*(datacols+4) 0x955cc78
*(datacols+5) 0x9576938

Please note that the first block allocated starts at 0x94f5978 which has already been allocated for the malloc call before the for loop. The compiler for some reason seem to reserve only 12 bytes instead of 24.

Can some one explain / help. I'm wondering what I'm doing wrong.

I'm running gcc4.0.1 on Fedora Core 4
I'm able to duplicate the same error on gcc running in Cygwin.
Thanks
Senthil

PS: Please note that colstofindcount and colslen are being calculated at runtime, and I have just put in the numbers for this example.

http://www.programmersheaven.com/c/M...ing=A9999F0001

Quote:

---

The compiler for some reason seem to reserve only 12 bytes instead of 24.

---

0x94f5968 - 0x94f598c = 0x24 = 36 (decimal)

0x94f5968 + 24 (decimal) = 0x94F5980.

I don't have the same problem:

Code:

---

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
  int i;
  int nrows = 6, ncols = 26416;
  long int **data;

  data = malloc(sizeof(long int *) * nrows);
  printf("data = %p\n\n", data);

  for(i = 0;i < nrows;++i)
  {
    *(data + i) = malloc(sizeof(long int) * ncols);
    printf("*(data + %d) = %p\n", i, *(data + i));
  }

  return 0;
}

---

Code:

---

itsme@itsme:~/C$ ./multimem
data = 0x804a050

*(data + 0) = 0x804a070
*(data + 1) = 0x8063d38
*(data + 2) = 0x807da00
*(data + 3) = 0x80976c8
*(data + 4) = 0x80b1390
*(data + 5) = 0x80cb058

---

0x804a070 - 0x804a050 = 0x20 = 32 (decimal)

That's using gcc-3.4.4

Firstly, thx everyone. Dave, I did the additon in hex mode when posting. Sorry ! about that.

The problem is with the different variables I have used in the first malloc and the following for loop

Code:

---


datacols=(long int **)malloc(colstofindcount*sizeof(long int *));
for(i=0;i<noofcolstofind;i++){

---

Those two variables, should be the same, but they are not and have different values. During runtime, colstofindcount is zero, malloc just returns a valid pointer, but zero bytes are allocated.

I'm sorry about the line which says

Code:

---

colstofindcount=6;

---

I just added that during the original post, to make the question clearer, but I was not paying enough attention.

Thanks again for all the help and suggestions

Senthil