C ++ 17使用类模板参数在一个包含函数返回值的类型上的Deduction指南

Question

I designed a object that takes in a function and its parameters and holds the function's return value inside the object to be retrieved later.

My goal here is to create a deduction guide for this object that'll allow me to omit the return type of the function in the object's constructor.

#include <utility>

template <typename Ret> class footure {
  public:
    template <typename Function, typename... Args>
    explicit footure(Function &&fun, Args &&... args) {
        // ...
        value_ = fun(std::forward<Args>(args)...);
        // ...
    }

    Ret get() { return value_; }

  private:
    Ret value_;
};

int add(const int a, const int b) { return a + b; }

int main() {
    auto f = footure<int>(add, 2, 3); // Remove <int>
    auto r = f.get();
}

I've looked up resources such as this PR to try and figure this out, but I could not come up with a solution.

Answer 1

You seem to look for std::invoke_result .

template <typename Function, typename... Args>
explicit footure(Function&&, Args&&...) -> footure<std::invoke_result_t<Function&&, Args&&...>>;

Don't forget to add the header <type_traits> .

Answer 2

The class must know about the Function and Args at compile time. Only then can you deduce the type of _value like so:

#include <utility>
#include <type_traits>

template <typename Function, typename... Args> class footure {
  public:

    explicit footure(Function &&fun, Args &&... args) {
        // ...
        value_ = fun(std::forward<Args>(args)...);
        // ...
    }

    auto get() { return value_; }


  private:
     typename std::result_of<Function&(Args...)>::type value_;
};

int add(int a, int b) { return a +b; }

int main() {
    auto f = footure(add, 1, 2); // Remove <int>
    auto r = f.get();
}