Constexpr function is not constexpr when template is used

Question

The following code compiles fine:

struct A {
    int i;
    constexpr A() : i(1) { }
    constexpr A(const A& that) : i(1) { }
};
constexpr auto func() {
    std::array<A, 3> result = {};
    return result;
}

However, if we add a template type parameter T to A,

template<typename T> struct A {
    int i;
    constexpr A() : i(1) { }
    constexpr A(const A<T>& that) : i(1) { }
};
constexpr auto func() {
    std::array<A<int>, 3> result = {};
    return result;
}

the compiler errors "constexpr function 'func' cannot result in a constant expression".

How is this possible?

Which compiler? On Coliru, it compiles fine.

davidhigh
– davidhigh

2019-02-02 21:58:06 +00:00
Commented Feb 2, 2019 at 21:58 — davidhigh
– davidhigh, Commented Feb 2, 2019 at 21:58
Compiler: Microsoft visual C++ 14.16.27023

Carucel
– Carucel

2019-02-02 21:58:38 +00:00
Commented Feb 2, 2019 at 21:58 — Carucel
– Carucel, Commented Feb 2, 2019 at 21:58

davidhigh · Accepted Answer · 2019-02-02 22:07:36Z

2

Yes, MSVC had (or still has) some problems with the implementation of C++14/17 features, and that obviously also applies to constexpr. With Visual Studio 2017 15.9, however, the following slight modification works for me (whereas the version in the OP also gives an error):

template<typename T> struct A {
    int i;
    constexpr A() : i(1) { }
    constexpr A(const A<T>& that) : i(1) { }
};
constexpr auto func() {
    return std::array<A<int>, 3>{};
}

answered Feb 2, 2019 at 22:07

davidhigh

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2 Comments

Carucel Over a year ago

Thank you. Is this a bug in the MSVC compiler?

davidhigh Over a year ago

Yes, it's a bug. It's probably something related to the copy constructor or to copy(-list) initialization.

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