static assert that template typename T is NOT complete?

c++, incomplete-type, static-assert, templates

Solution

Here is a function using expression SFINAE based on chris proposal which allows checking whether a type is complete yet. My adoption needs no includes, errors-out when the required argument is missing (hiding the argument was not possible) and is suitable for C++11 onward.

template<typename T>
constexpr auto is_complete(int=0) -> decltype(!sizeof(T)) {
    return true;   
}

template<typename T>
constexpr bool is_complete(...) {return false;}

And a test-suite:

struct S;

bool xyz() {return is_complete<S>(0);}

struct S{};

#include <iostream>
int main() {
    std::cout << is_complete<int>(0) << '\n';
    std::cout << xyz() << '\n';
    std::cout << is_complete<S>(0);
}

Output:

1
0
1

See live on coliru

Problem

Is there a way to static_assert that a type T is Not complete at that point in a header? The idea is to have a compile error if someone adds #includes down the road in places they should not be. related: How to write `is_complete` template? Using that link's answer, ``` namespace { template<class T, int discriminator> struct is_complete { static T & getT(); static char (& pass(T))[2]; static char pass(...); static const bool value = sizeof(pass(getT()))==2; }; } #define IS_COMPLETE(X) is_complete<X,__COUNTER__>::value class GType; static_assert(!IS_COMPLETE(GType),"no cheating!"); ``` unfortunately this gives "invalid use of incomlete type" error, d'oh. Is there a way to assert on the negation?

Original source

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