Is comparing two void pointers to different objects defined in C++? - c++

Question

Inspired by this answer about dynamic cast to void*:
...
bool eqdc(B* b1, B *b2) {
return dynamic_cast<void*>(b1) == dynamic_cast<void*>(b2);
}
...
int main() {
DD *dd = new DD();
D1 *d1 = dynamic_cast<D1*>(dd);
D2 *d2 = dynamic_cast<D2*>(dd);
... eqdc(d1, d2) ...
I am wondering if it is fully defined behaviour in C++ (according to the 03 or 11 standard) to compare two void pointers for (in)equality that point to valid, but different objects.
More generally, but possibly not as relevant, is comparing (==or !=) two values of type void* always defined, or is it required that they hold a pointer to a valid object/memory area?

Answer 1

C says:
Two pointers compare equal if and only if both are null pointers, both are pointers to the
same object (including a pointer to an object and a subobject at its beginning) or function,
both are pointers to one past the last element of the same array object, or one is a pointer
to one past the end of one array object and the other is a pointer to the start of a different
array object that happens to immediately follow the first array object in the address
space.
C++ says:
Two pointers of the same type compare equal if
and only if they are both null, both point to the same function, or both represent the same address.
Hence it would mean that:
a)
it is fully defined behaviour in C++ (according to the 03 or 11 standard) to compare two void pointers for (in)equality that point to valid, but different objects.
So yes, in both C and C++. You can compare them and in this case they shall compare as true iff they point to the same object. That's simple.
b)
is comparing (==or !=) two values of type void* always defined, or is it required that they hold a pointer to a valid object/memory area?
Again, the comparison is well-defined (standard says "if and only if" so every comparison of two pointers is well-defined). But then...
C++ talks in terms of "address", so I think this means that the standard requires this to work "as we'd expect",
C, however, requires both the pointers to be either null, or point to an object or function, or one element past an array object. This, if my reading skills aren't off, means that if on a given platform you have two pointers with the same value, but not pointing to a valid object (e.g. misaligned), comparing them shall be well-defined and yield false.
This is surprising!
Indeed that's not how GCC works:
int main() {
void* a = (void*)1; // misaligned, can't point to a valid object
void* b = a;
printf((a == b) ? "equal" : "not equal");
return 0;
}
result:
equal
Maybe it's UB in C to have a pointer which isn't a null pointer and doesn't point to an object, subobject or one past the last object in an array? Hm... This was my guess, but then we have that:
An integer may be converted to anypointer type. Except as previously specified, the
result is implementation-defined, might not be correctly aligned, might not point to an
entity of the referenced type, and might be a trap representation.
So I can only interpret it that the above program is well-defined and the C standard expects it to print "not equal", while GCC doesn't really obey the standard but gives a more intuitive result.

Answer 2

C++11, 5.10/1:
Pointers of the same type (after pointer conversions) can be compared
for equality. Two pointers of the same type compare equal if and only
if they are both null, both point to the same function, or both
represent the same address
So yes, the specific comparison is OK.
In general it is undefined behavior to attempt to create a pointer value that isn't a valid address - for example using pointer arithmetic to go before the beginning or after the one-after-the-end of an array - let alone use them. The result of stuff like (void*)23 is implementation-defined, so barring specific permission from the implementation it is in effect undefined behavior to compare those too, since the implementation might define that the result is a trap value of void*.