是否可以在C ++中使用函数返回类型作为参数来声明另一个函数？

Question

I want something like this:

std::tuple<int, bool, double> MyFunction_1 (void);

void MyFunction_2 (decltype (MyFunction_1) &params);

Obviously, in this example a code pointer to function would be passed.

I want to have the equivalent of this:

void MyFunction_2 (std::tuple<int, bool, double>  &params);

Is it possible to do so?

Answer 1

decltype (MyFunction_1) will give you the type of MyFunction_1 (ie the function type std::tuple<int, bool, double> () ), you need to emulate a function calling ¹ (via adding () ) to get the return type (ie std::tuple<int, bool, double> ), eg

void MyFunction_2 (decltype (MyFunction_1()) &params);
//                                       ^^

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^{1 The expression is evaluated at compile-time, the function won't be called at run-time actually.}

Answer 2

The type of MyFunction_1 is not std::tuple<int, bool, double> - informally you can think of it as a function pointer. In fact &MyFunction_1 decays to itself and is certainly a pointer.

So decltype(MyFunction_1) is not what you want.

The solution is to write decltype(MyFunction_1()) . The type of MyFunction_1() is std::tuple<int, bool, double> . Note that this doesn't actually call the function; it's rather like sizeof in that respect.

Answer 3

using RetType = std::invoke_result<decltype(f)>::type;

玩它： https ： //gcc.godbolt.org/z/AE7lj_