Casting between primitive type pointers

Question

Is the following well-defined:

char* charPtr = new char[42];
int* intPtr = (int*)charPtr;

charPtr++;
intPtr = (int*) charPtr;

The in

Answer 1

In general, the result is unspecified (5.2.10p7) if the alignment requirements of int are greater than those of char, which they usually will be. The result will be a valid value of the type int * so it can be e.g. printed as a pointer with operator<< or converted to intptr_t.

Because the result has an unspecified value, unless specified by the implementation it is undefined behaviour to indirect it and perform lvalue-to-rvalue conversion on the resulting int lvalue (except in unevaluated contexts). Converting back to char * will not necessarily round-trip.

However, if the original char * was itself the result of a cast from int *, then the cast to int * counts as the second half of a round trip; in that case, the cast is defined.

In particular, in the case above where the char * was the result of a new[] expression, we are guaranteed (5.3.4p10) that the char * pointer is appropriately aligned for int, as long as sizeof(int) <= 42. Because the new[] expression obtains its storage from an allocation function, 3.7.4.1p2 applies; the void * pointer can be converted to a pointer of any complete object type with a fundamental alignment requirement and then used to access the object [...] which strongly implies, along with the note to 5.3.4p10, that the same holds for the char * pointer returned by the new[] expression. In this case the int * is a pointer to an uninitialised int object, so performing lvalue-to-rvalue conversion on its indirection is undefined (3.8p6), but assigning to its indirection is fully defined. The int object is in the storage allocated (3.7.4.1p2) so converting the int * back to char * will yield the original value per 1.8p6. This does not hold for the incremented char * pointer as unless sizeof(int) == 1 it is not the address of an int object.