Constexpr静态成员函数用法

Question

Consider the following example code:

#include <array>

struct MyClass
{
  size_t value = 0;

  constexpr static size_t size() noexcept
  {
    return 3;
  }
};

template <size_t N>
void DoIt()
{
  MyClass h;
  std::array<int, h.size()> arr;
}

int main()
{
  DoIt<1>();
}

When I try to compile this with GCC 7.3.0, I get an error about h not being usable in a non-constexpr context:

cexpr.cpp: In function ‘void DoIt()’:
cexpr.cpp:17:26: error: the value of ‘h’ is not usable in a constant expression
   std::array<int, h.size()> arr;
                          ^
cexpr.cpp:16:11: note: ‘h’ was not declared ‘constexpr’
   MyClass h;
           ^
cexpr.cpp:17:27: error: the value of ‘h’ is not usable in a constant expression
   std::array<int, h.size()> arr;
                           ^
cexpr.cpp:16:11: note: ‘h’ was not declared ‘constexpr’
   MyClass h;
           ^

However, when I try compiling the exact same code in Clang 6.0.0, it compiles without any errors. Additionally, when I modify the code to not be inside the templated DoIt() function, GCC compiles this just fine:

#include <array>

struct MyClass
{
  size_t value = 0;

  constexpr static size_t size() noexcept
  {
    return 3;
  }
};

int main()
{
  MyClass h;
  // this compiles just fine in Clang and GCC
  std::array<int, h.size()> arr;
}

I already know how to fix the first code so it compiles on GCC using decltype , but I'm curious to know why doesn't the first piece of code compile with GCC? Is this just a bug in GCC, or is there something I don't understand about using constexpr static member functions?

Answer 1

Looks like a bug to me.

The type and meaning of the expression h.size() is defined by [expr.ref] "Class member access":

[expr.post]/3

Abbreviating postfix-expression.id-expression as E1.E2 , E1 is called the object expression. [...]

and

[expr.post]/6.3.1

If E2 is a (possibly overloaded) member function, function overload resolution is used to determine whether E1.E2 refers to a static or a non-static member function.

• (6.3.1) If it refers to a static member function and the type of E2 is “function of parameter-type-list returning T ”, then E1.E2 is an lvalue; the expression designates the static member function . The type of E1.E2 is the same type as that of E2 , namely “function of parameter-type-list returning T ”.

This means h.size has the same type as ::MyClass::size and is evaluated as such, regardless of the fact that h is constexpr or not .

h.size() is then a call to a constexpr function and is a core constant expression according to [expr.const]/4 .