How to interpret section 6.3.2.3 part 7 of the C11 standard?

Question

In the context of another question there was some discussion on whether it was allowed (i.e. would or would not introduce implementation defined or undefined behavior) to cast int** to void** and subsequently assign a value to the dereferenced void*. This brings me to my question on the interpretation of the C11 standard

6.2.5 (28) A pointer to void shall have the same representation and alignment requirements as a pointer to a character type. ...
6.3.2.3 (1) A pointer to void may be converted to or from a pointer to any object type. A pointer to any object type may be converted to a pointer to void and back again; the result shall compare equal to the original pointer.
6.3.2.3 (7) ... When a pointer to an object is converted to a pointer to a character type, the result points to the lowest addressed byte of the object. ...

My question is whether this

int* intptr = NULL;
void* dvoidptr = &intptr; /* 6.3.2.3 (1) */
*(void**)dvoidptr = malloc(sizeof *intptr); /* using 6.3.2.3 (1) */

conforms with the standard or not? It seems strange to me, but I cannot find a conclusive line of argument why not. void* to void** is guaranteed by 6.3.2.3 and 6.2.5 together with 6.3.2.3 help with the alignment.

Answer 1

Code is not valid.

See the emphasis:

6.3.2.3 (1) A pointer to void may be converted to or from a pointer to any object type. A pointer to any object type may be converted to a pointer to void and back again; the result shall compare equal to the original pointer.

You are converting int** to void* (fine), and then to different type void** (not ok). C gives no quarantee that you can safely convert void* to anything other than the original type (aside from converting to char *)

In addition, *(void**)dvoidptr results in void* when in reality there is int*. What if, for example, sizeof(void*) == 2 and sizeof(int*) == 1? You can convert void* pointer to other type, but you cannot reinterpret it directly as other type.