How does free know how much to free?

Question

In C programming, you can pass any kind of pointer you like as an argument to free, how does it know the size of the allocated memory to free? Whenever I pass a pointer to some function, I have to also pass the size (ie an array of 10 elements needs to receive 10 as a parameter to know the size of the array), but I do not have to pass the size to the free function. Why not, and can I use this same technique in my own functions to save me from needing to cart around the extra variable of the array's length?

Answer 1

You cannot pass any kind of pointer to free(). You can only pass the pointer that was earlier allocated via

malloc/calloc/strdup

Besides, you can also do free(NULL) and it would be ignored.

The memory block that is returned by malloc/calloc/strdup is of slightly more size than requested by the user. There is a small control header preceding the user pointer and that contains control information about the block and possibly other elements in the heap.