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

--
This message has been scanned for viruses and
dangerous content by MailScanner, and is
believed to be clean.

_______________________________________________
freebsd-questions@freebsd.org mailing list
http://lists.freebsd.org/mailman/listinfo/freebsd-questions
To unsubscribe, send any mail to "[EMAIL PROTECTED]"

Reply via email to