vs 2015 constexpr 变量非常量但在 vs 2019 上很好？

Question

I have a simple constexpr class (using c++17)

struct foo {
   float x, y;

   // ill formed constexpr in vs 2015
   foo() {}

   constexpr foo(float x, float y) : x(x), y(y) {
   }

};

constexpr auto bar() {
    return foo(4.0, 5.0);
}

int main() {
    auto f = bar();
}

This is illformed bc of the non constexpr default constructor on 2015. But vs 2019 doesnt report a problem.

According to this reference page "all selected ctors" must be constexpr. I assume the word "selected" means "used" and even though default ctor is unused, 2015 wont eval into a constexpr until I make the default ctor constexpr.

Other specs are worded differently but I could not make clear of them either. Also most constexpr examples use the keyword everywhere. An example of selective constexpr use would be nice.

Latest GNU ans Clang behave like vs 2019 but Im not convinced that this was an oversight or bug in vs 2015.

So what is it? Should all ctors be constexpr or is visual studio 2019 correct here?

Answer 1

There has never been a requirement that all constructors of a literal type must be constexpr . The requirement has only been that all functions which get invoked in a constant expression context must be able to be constant expressions (and thus must be declared constexpr ).

Note that older versions of Visual Studio did not do a good job of implementing constexpr correctly.