malloc'd内存和sigsegv

Question

help me in understanding the malloc behaviour.. my code is as follows::

    int main()
    {   
    int *ptr=NULL;
    ptr=(int *)malloc(1);
    //check for malloc
    *ptr=1000;

    printf("address of ptr is %p and value of ptr is %d\n",ptr,*ptr);
    return 0;
    }

the above program works fine(runs without error)...how?? as I have supplied a value of 1000 in 1 byte only!!

Am I overwriting the next memory addresss in heap? if yes, then why not sigsgev is there?

Answer 1

Many implementations of malloc will allocate at a certain "resolution" for efficiency.

That means that, even though you asked for one byte, you may well have gotten 16 or 32.

However, it's not something you can rely on since it's undefined behaviour.

Undefined behaviour means that anything can happen, including the whole thing working despite the problematic code :-)

Answer 2

Using a debug heap you will definitely get a crash or some other notification when you freed the memory (but you didn't call free).

Segmentation faults are for page-level access violations, and a memory page is usually on the order of 4k, so an overrun by 3 bytes isn't likely to be detected until some finer grained check detects it or some other part of your code crashes because you overwrote some memory with 'garbage'