memory allocation with malloc

[Derek Ragona]

At 01:16 AM 8/5/2008, Shyamal Shukla wrote:
>Hi All,
>
>      I am trying to validate my understanding of how malloc works by means
>of the below C program which tries to corrupt essential information
>maintained by malloc for free() operation.
>
>The program allocates 4, 12 byte blocks (internally 16 bytes are allocated
>for each 12 byte block). Hence the total allocated space was 48 bytes.
>
>As malloc maintains the (length of allocated block + 1), 4 bytes before the
>returned pointer (from malloc), I have manipulated this length for the first
>block and set it to 49 with the goal that a single free shall release all
>these 4 blocks and a subsequent malloc of 15 bytes shall be from the address
>of first block.
>
>However, this does not happen. Can someone please correct my understanding
>and provide me with a reference to the working of malloc() and free()?
>
>#include<stdio.h>
>
>int main(void)
>{
>     char * ptr,* ptr1, *ptr2, * ptr3, * ptr4;
>     int * i;
>     int n,q,p;
>     int loop = 0;
>
>     ptr1 = (char *)malloc(12);
>     i = (int *)(ptr1 - 4);
>     printf("\n ptr1 = %p,%d \n",ptr1,*i);
>     printf("\n %d:%d:%d:%d\n",ptr1[-4],ptr1[-3],ptr1[-2],ptr1[-1]);
>     printf("\n %d:%d:%d:%d\n",ptr1[0],ptr1[1],ptr1[2],ptr1[3]);
>     printf("\n %d:%d:%d:%d\n",ptr1[4],ptr1[5],ptr1[6],ptr1[7]);
>     printf("\n %d:%d:%d:%d\n",ptr1[8],ptr1[9],ptr1[10],ptr1[11]);
>     *i = 49;
>
>     ptr2 = (char *)malloc(12);
>     i = (int *)(ptr2 - 4);
>     printf("\n ptr2 = %p,%d \n",ptr2,*i);
>     printf("\n %d:%d:%d:%d\n",ptr2[-4],ptr2[-3],ptr2[-2],ptr2[-1]);
>
>     ptr3 = (char *)malloc(12);
>     i = (int *)(ptr3 - 4);
>     printf("\n ptr3 = %p,%d \n",ptr3,*i);
>     printf("\n %d:%d:%d:%d\n",ptr3[-4],ptr3[-3],ptr3[-2],ptr3[-1]);
>
>     ptr4 = (char *)malloc(12);
>     i = (int *)(ptr4 - 4);
>     printf("\n ptr4 = %p,%d \n",ptr4,*i);
>     printf("\n %d:%d:%d:%d\n",ptr4[-4],ptr4[-3],ptr4[-2],ptr4[-1]);
>
>     free(ptr1);
>     printf("\n ------------ANALYZE-------------\n");
>     printf("\n %d:%d:%d:%d\n",ptr1[-4],ptr1[-3],ptr1[-2],ptr1[-1]);
>     printf("\n %d:%d:%d:%d\n",ptr1[0],ptr1[1],ptr1[2],ptr1[3]);
>     printf("\n %d:%d:%d:%d\n",ptr1[4],ptr1[5],ptr1[6],ptr1[7]);
>     printf("\n %d:%d:%d:%d\n",ptr1[8],ptr1[9],ptr1[10],ptr1[11]);
>
>     ptr = (char *)malloc(15);
>     i = (int *)(ptr - 4);
>     printf("\n ptr = %p,%d \n",ptr,*i);
>     return;
>}
>
>
>Thanks and Regards,
>Shyamal
>
>

I'm not quite sure what it is you want to accomplish with this 
program.  However, malloc and free work on the program's given data 
area.  This data area can be increased should there be a need for more 
memory.

You should NEVER assume that memory blocks are contiguous.  There are many 
reasons why they would not be contiguous among them compiler 
optimizations.  If you really want to delve into how a program is executed, 
have the compiler output the assembler code and look at that.  The 
assembler code will show exactly how and where the variables are 
allocated.  With such small amount of data used in your program, it is 
possible the variables are all just on the stack.


You may want to check out the brk and sbrk man pages as they will give you 
some information into how memory management was originally done as these 
functions are lower-level than malloc and free.

         -Derek

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