Malloc，free和多个指针，它如何工作？

Question

I have multiple pointers pointing to dynamically allocated memory which was assigned using single malloc.

int *p1, *p2;
p1 = (int*)malloc(sizeof(int) * 10);
p2 = &p1[5];
free(p2);
p1[5] = 3;
p1[6] = 5;

Now my question is what does 'free(p2)' statement would do? Will it free memory ahead p1[5]? Is it safe to use memory ahead p1[5](ie p1[6], p1[7] , . .)

Answer 1

Freeing a pointer that wasn't malloc / calloc ed is undefined behavior. Some allocators such as glibc's can sometimes detect it and deterministically crash the program with a free(): invalid pointer error message, but technically you lose any and all guarantees about your program's behavior if you do it.

Answer 2

You cannot predict the behavior of this function, as it is undefined behavior

From the reference:

Deallocates the space previously allocated by malloc() , calloc() , aligned_alloc , (since C11) or realloc() .

If ptr is a null pointer, the function does nothing.

The behavior is undefined if the value of ptr does not equal a value returned earlier by malloc() , calloc() , realloc() , or aligned_alloc() (since C11).

— free , C++ Reference

Emphasis mine: your use of free in this context would involve freeing from a pointer that was not obtained from the use of any of those functions; it was obtained by transforming the pointer that was obtained from malloc , and thus is not valid.

My best guess at what might happen is a segmentation fault; but that's up to your compiler, and not something you or I can guarantee.

So do not do this.