What is const void?

The description of std::is_void states that:

Provides the member constant value that is equal to true, if T is the type void, const void, volatile void, or const volatile void.

Then what could be const void, or a volatile void ?

This answer states that const void return type would be invalid (however compiles on VC++ 2015)

const void foo() { }

If by standard, const void is invalid (VC being wrong) - then what is const void?

Solution 1:

const void is a type which you can form a pointer to. It's similar to a normal void pointer, but conversions work differently. For example, a const int* cannot be implicitly converted to a void*, but it can be implicitly converted to a const void*. Likewise, if you have a const void* you cannot static_cast it to an int*, but you can static_cast it to a const int*.

const int i = 10;
void* vp = &i;                           // error
const void* cvp = &i;                    // ok
auto ip = static_cast<int*>(cvp);        // error
auto cip = static_cast<const int*>(cvp); // ok

Solution 2:

As void, const void is a void type. However, if const void is a return type, the const is meaningless (albeit legal!), because [expr]/6:

If a prvalue initially has the type “cv T”, where T is a cv-unqualified non-class, non-array type, the type of the expression is adjusted to T prior to any further analysis.

However, it is a valid type itself and occurs in e.g. C-standard library functions, where it's used to ensure const-correctness of argument pointers: int const* cannot be converted to void*, but void const*.

Solution 3:

Types can be the result of templates; a template might state const T, and be instantiated with T as void.

The linked answer is misled, or rather, limited in view in that it regards the special case of a non-template type, and even then const void might be meaningless, but it is valid code.