Why const literal class is not a constexpr even if initialized from a constexpr?[code] class test{ public: constexpr test(){} constexpr int operator+(const test& rhs) ...
Why do I have to surround parentheses in this fold expression?[code] template<typename... Args> void printVariadic(Args... args) { (... , (std::cout << args << '...
It is known that we can't initialize rvalue reference with lvalue, but why can function name?Let me confirm your answer first. The link you gave me is "No, because rvalue references can be init...
It is known that we can't initialize rvalue reference with lvalue, but why can function name?@mbozzi: I disagree. Your link does nothing but repeat my code and say "look, it works". I don't thi...
It is known that we can't initialize rvalue reference with lvalue, but why can function name?It is said that we can't initialize an rvalue reference with an lvalue. For example: [code]int i=0; ...
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