Purpose of perfect forwarding for Callable argument in invocation expression?

孤人 提交于 2019-11-28 01:51:43

For the same purpose as for arguments: so when Func::operator() is a ref-qualified:

struct Functor
{
    void operator ()() const &  { std::cout << "lvalue functor\n"; }
    void operator ()() const && { std::cout << "rvalue functor\n"; }
};

Demo

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