int main ()
{
int * b;
b = (int*) malloc (1);
*b=110000;
free (b);
return 0;
}
Why does heap corruption happen at free (b);?
IMO, heap corruption already happens at *b=110000;.
Answer
malloc()'s argument is the number of bytes to allocate. You need to use:
b = (int*) malloc(sizeof(int));
You've allocated too small a block, and then written more bytes to it than you've allocated, which overwrites bookkeeping information next to the block, corrupting the heap.