What does '->' mean after a function declaration in C++ 11?

Question

What is the syntax used here with '-> decltype' after the operator() signature and what is it for?

template<>
struct less<void>
{   // transparent functor for operator<
typedef int is_transparent;

template<class _Ty1,
    class _Ty2>
    constexpr auto operator()(_Ty1&& _Left, _Ty2&& _Right) const
    -> decltype(static_cast<_Ty1&&>(_Left)
        < static_cast<_Ty2&&>(_Right))
    {   // transparently apply operator< to operands
    return (static_cast<_Ty1&&>(_Left)
        < static_cast<_Ty2&&>(_Right));
    }
};

this is the code from C:\Program Files (x86)\Microsoft Visual Studio\2017\Community\VC\Tools\MSVC\14.15.26726\include\xstddef line 276.

Why are the following two lines duplicated?

(static_cast<_Ty1&&>(_Left)
        < static_cast<_Ty2&&>(_Right))

https://stackoverflow.com/questions/22514855/arrow-operator-in-function-heading The first answer should help — Lee Katsnelson, Sep 06 '18 at 11:55

score 0 · Answer 1 · answered Sep 06 '18 at 12:00

0

->decltype(static_cast<_Ty1&&>(_Left) returns the type of static_cast<_Ty1&&>(_Left) and declare it as the return type of the function.

E.G.

auto function(int x ) -> int and int function(int x ) are the same

answered Sep 06 '18 at 12:00

Simo

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What does '->' mean after a function declaration in C++ 11?

1 Answers1