What is the difference between a void* and *(void**)?

Question

I was programming an `aligned_malloc` implementation in C++ and mainly use the following code, which can be found in various places on the Interwebs, including SO: ``` void * aligned_malloc(size_t size, int align) { if (align < 0) { return NULL; } void *ptr; void *p = malloc(size + align - 1 + sizeof(void*)); if (p != NULL) { ptr = (void*) (((ptrdiff_t)p + sizeof(void*) + align -1) & ~(align-1)); *((void**)((ptrdiff_t)ptr - sizeof(void*))) = p; return ptr; } return NULL; } void aligned_free(void *p) { void *ptr = *((void**)((ptrdiff_t)p - sizeof(void*))); free(ptr); return; } ``` I get the `*(void**)` in the `aligned_malloc`. My misunderstanding is with the cast in the `aligned_free`. Since we just want the value and don't have to assign to it why not use ``` void *ptr = ((void*)((ptrdiff_t)p - sizeof(void*))); ``` instead of ``` void *ptr = *((void**)((ptrdiff_t)p - sizeof(void*))); ``` I thought it was the same but when I tried the first one with an alignment of 64 it gave me an err…

Accepted Answer

Let us simplify the code:

```
int x = 7;
void * p = &x;
void * q = &p;
```

And now let us draw a diagram:

```
  +---+
x | 7 |
  +---+
    ^ 
    |
  +-|-+
p | * |
  +---+
    ^ 
    |
  +-|-+
q | * |
  +---+
```

Do you see how `p` and `q` are not equal? If you have `q`, and you want to get to `p`, you have to dereference `q`. But since a `void*` cannot be dereferenced, you have to promise the type system that it will find another pointer after the dereference, hence the cast to `void**` before the dereference.

```
assert(q != p);
assert(*(void**)q == p);
```

What is the difference between a void* and *(void**)?

Solution

Problem