Why can't I use google test's ASSERT_TRUE for my SFINAE check?

Question

I have a simple SFINAE check for the existence of a static member function using C++17's std::void_t:

template <class T, class X = void>
struct has_bar {
    static const bool value = false;
};
template <class T>
struct has_bar<T, std::void_t<decltype(T::bar(std::declval<int&>()))>> {
    static const bool value = true;
};

The class Foo obviously has a bar member function taking an int& as its only argument, the class NoBar has no such member function:

struct Foo
{
    static void bar(int& value)
    {
        value = 42;
    }
};

struct NoBar{};

Using has_bar in a static_assert works as expected, but using it in an ASSERT_TRUE statement yields a compiler error:

TEST(SFINAE, has_bar)
{
    static_assert(!has_bar<NoBar>::value);
    static_assert(has_bar<Foo>::value);
    ASSERT_FALSE(has_bar<NoBar>::value);

    //why does this not work??
    ASSERT_TRUE(has_bar<Foo>::value);
}

int main(int argc, char** argv)
{
    ::testing::InitGoogleTest(&argc, argv);
    return RUN_ALL_TESTS();
}

ASM generation compiler returned: 0
/opt/compiler-explorer/gcc-10.2.0/bin/../lib/gcc/x86_64-linux-gnu/10.2.0/../../../../x86_64-linux-gnu/bin/ld: /tmp/ccfSP5vL.o: in function `SFINAE_has_bar_Test::TestBody()':
/home/ce/<source>:42: undefined reference to `has_bar<Foo, void>::value'
collect2: error: ld returned 1 exit status
Execution build compiler returned: 1

Is this a google test bug or am I missing something here?

Live Code C++17: https://godbolt.org/z/fj8z6b
Live Code C++14: https://godbolt.org/z/n7TvqW

Does it pass if you change the check to `ASSERT_TRUE((bool)has_bar::value);` ? I know it seems pointless, but humor me please. — StoryTeller - Unslander Monica, Feb 23 '21 at 20:38
It's pretty much the same explanation as presented in this answer (https://stackoverflow.com/a/272982/817643). Seems the macro hides some sort of reference binding. — StoryTeller - Unslander Monica, Feb 23 '21 at 20:42
Thanks, that makes sense! I would never have found this so quickly... — joergbrech, Feb 23 '21 at 20:49
Happy to help. Unlike the answer I linked though, I wouldn't recommend a macro. A possible solution is not to define `value` yourself. Instead your trait could inherit from `std::false_type` and `std::true_type`. The standard library should be handling the ODR correctly. — StoryTeller - Unslander Monica, Feb 23 '21 at 20:55
HInt: derive from std::false_type or std::true_type instead of defining static member value (both types from type_traits) — PiotrNycz, Feb 23 '21 at 20:59

score 2 · Accepted Answer · answered Feb 23 '21 at 20:59

This error occurs because has_bar::value has been declared but not defined. Somehow, the ASSERT_TRUE macro has odr-used this object. It's not necessary to understand exactly what ASSERT_TRUE is doing; you should always define your static class members.

The easiest way to do this is to inherit from std::false_type and std::true_type instead of declaring a static member named value. The standard library already contains a definition for the inherited value member, so you don't have to provide your own.

In C++17, note that static constexpr bool value = false; is a definition by itself (and is implicitly inline). But this isn't the case if the member is merely const and not constexpr, so you see the issue in both C++14 and C++17 mode.

Why can't I use google test's ASSERT_TRUE for my SFINAE check?

1 Answers1