CODEWITHSUNDEEP
c++visual-studio-2015c++14
Adrian
Adrian

Reputation: 10911

Deduction failure when using a constexpr function?

So I was doing further testing based on this question, and I'm still a little unclear about how type deduction works.

If I use the following:

template<typename T, std::enable_if_t<std::is_same<T,int>::value, int> = 0>
inline auto fnx(T) -> int
{
    return 0;
}

template<typename T, std::enable_if_t<std::is_same<T, float>::value, int> = 0>
inline auto fnx(T) -> int
{
    return 0;
}

inline void fn()
{
    fnx(1);
    fnx(1.f);
}

I get no compilation errors. But when I do this:

template <bool TRUTH>
constexpr bool value() { return TRUTH; }

template<typename T, std::enable_if_t<value<std::is_same<T,int>::value>(), int> = 0>
inline auto fnx(T) -> int
{
    return 0;
}

template<typename T, std::enable_if_t<value<std::is_same<T, float>::value>(), int> = 0>
inline auto fnx(T) -> int
{
    return 0;
}

inline void fn()
{
    fnx(1);
    fnx(1.f);
}

The type deduction fails. What is happening here? Why is going through a constexpr function making it not valid? Or is this a problem with my c++ compiler? I'm using VC++2015.

The errors are:

error C2672: 'detail::fnx': no matching overloaded function found
error C2783: 'int detail::fnx(T)': could not deduce template argument for '__formal'
note: see declaration of 'detail::fnx'

Upvotes: 5

Views: 592

Answers (2)

Guillaume Racicot
Guillaume Racicot

Reputation: 41750

Your code is fine and completely valid. You must however not use constexpr function to do SFINAE with visual studio 2015. You can read more about expression SFINAE in MSVC here: Expression SFINAE improvements in VS 2017 RC

Upgrading to 2017 will solve your issue.

Upvotes: 2

AndyG
AndyG

Reputation: 41092

This is a bug in your version of Visual Studio. Try upgrading to 2017.

I could not reproduce with 19.00.23506

Note that gcc 7.0.1 and Clang 5.0.0 had no issues.

Upvotes: 2

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