What is Allowed in a constexpr Function?

Question

constexpr functions are not supposed to contain:

A definition of a variable of non-literal type

But in this answer a lambda is d

Answer 1

It's virtually guaranteed that if there's a discrepancy gcc has the correct behavior, because Visual Studio 2015 doesn't support c++14's extension of constexpr: https://msdn.microsoft.com/en-us/library/hh567368.aspx#C-14-Core-Language-Features

C++11 constexpr functions

The function body can only contain:

• null statements (plain semicolons)
• static_assert declarations
• typedef declarations and alias declarations that do not define classes or enumerations
• using declarations
• using directives
• exactly one return statement

So c++11 cannot tolerate the definition of decltype(div(T{}, T{})) x{}. It would however be acceptable to roll the ternary suggested here in a constexpr function to achieve the same results:

template <typename T>
constexpr auto make_div(const T quot, const T rem)
{
    using foo = decltype(div(T{}, T{}));
                         
    return foo{1, 0}.quot != 0 ? foo{quot, rem} : foo{rem, quot};
}

Live Example

C++14 constexpr functions

The function body may contain anything but:

• an asm declaration
• a goto statement
• a statement with a label other than case and default
• a try-block
• a definition of a variable of non-literal type
• a definition of a variable of static or thread storage duration
• a definition of a variable for which no initialization is performed

Where a "Literal Type" is defined here, specifically for objects though, they may be aggregate types with a trivial destructor. So div_t definitely qualifies. Thus c++14, and by extension gcc, can tolerate the definition of decltype(div(T{}, T{})) x{}.

C++17 constexpr functions

C++17 added support for closure types to the definition of "Literal Type", so I find it strange that both gcc and Visual Studio support the use of the lambda in the return statement. I guess that's either forward looking support or the compiler chose to inline the lambda. In either case I don't think that it qualifies as a c++14 constexpr function.

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