判断函数参数是否为函数

问题描述

如何确定函数的参数type是否为函数?我正在实现一个名为 Queue 的类，它接收一个参数.如果参数是函数，则存储该函数.

How can I determine whether a function's parameter type is a function? I'm implementing a class called Queue which receives a single parameter. If the parameter is a function, it stores the function.

代码如下:

template <class Type, typename Data>
class Queue {
    public:
        void Enqueue (Data& data) {
            if (typeid(data).name() == int) {
                intVector.push_back(data);
                order.push_back("int");
            }
            else if (typeid(data).name() == bool) {
                boolVector.push_back(data);
                order.push_back("bool");
            }
            else if (typeid().name() == string) {
              stringVector.push_back(data);
              order.push_back("string");
            }
            // This will continue for:
            // - double
            // - char
            // - function
        }

        auto Dequeue () {
            auto temp;
            switch (order.begin()) {
                case "int":
                    temp = intVector.begin();
                    intVector.erase(intVector.begin());
                    order.erase(order.begin());
                    return temp;
                // This will continue for:
                // - "string"
                // - "bool"
                // - "char"
                // - "double"
                // - "function"
                default:
                    cout << "An Error occurred while trying to Enqueue." << endl;
                    cout << "\tAddress: " << this << endl;
            }
        }

        auto Start () {
            // This function will run all of the processes...
        }
        Queue (Data& data) {
            if (typeid(Type).name() == int) {
                // Pseodo-code:
                // if (data.type == function) {
                //     Enqueue (data);
                // }
            }
        }
}

可以初始化:

Queue queue1 = new Queue <int> (func ()); // func () is a function.
Queue queue2 = new Queue <int> (var);     // var is a variable.

推荐答案

天哪.这是一个XY问题.

无论如何，在对 std::enable_if 进行了一番折腾后(这很有趣)，我意识到整个事情可以归结为:

Anyway, after messing around with std::enable_if for a bit (which was kinda fun), I realised that the whole thing can be boiled down to this:

#include <vector>
#include <string>
#include <any>
#include <iostream>
#include <functional>

void call_if_function (void (* f) ()) { f (); }
void call_if_function (std::function <void ()> f) { f (); }
void call_if_function (std::any x) { (void) x; }

template <class T>
class Queue
{
    public:
        void Enqueue (const T& data)
        {
//          std::cout << "Enqueueing " << data << "\n";
            v.push_back (data);
        }

        T Dequeue ()
        {
            T ret = v.front ();
//          std::cout << "Dequeueing " << ret << "\n";
            v.erase (v.begin ());
            call_if_function (ret);
            return ret;
        }

    private:
        std::vector <T> v;
};

而且，如果我正确理解 OP 的问题，那就是您所需要的全部.

And, if I understand the OP's problem right, that is all all you need.

测试程序:

void foo () { std::cout << "foo () called\n"; }
void bar (int x, int y) { std::cout << "bar () called, x = " << x << ", y = " << y << "\n"; }

int main ()
{
    // Queue of int's
    Queue <int> int_q;
    int_q.Enqueue (42);
    auto i = int_q.Dequeue ();
    std::cout << "int_q.Dequeue () returned " << i << "\n\n";

    // Queue of strings
    Queue <std::string> string_q;
    string_q.Enqueue ("Hello world");
    auto s = string_q.Dequeue ();
    std::cout << "string_q.Dequeue () returned " << s << "\n\n";

    // Call function with no parameters
    Queue <void (*)()> func_q;
    func_q.Enqueue (foo);
    auto f = func_q.Dequeue ();
    std::cout << "func_q.Dequeue () returned " << (void *) f << "\n";
    f ();

    // Call function with arbitrary parameters
    Queue <std::function <void ()>> func_qp;
    func_qp.Enqueue ([] () { bar (21, 99); });
    auto fp = func_qp.Dequeue ();
    fp ();
}

输出:

int_q.Dequeue () returned 42

string_q.Dequeue () returned Hello world

foo () called
func_q.Dequeue () returned 0x4026fd
foo () called

bar () called, x = 21, y = 99
bar () called, x = 21, y = 99

现场演示.

Moral:KISS，现在玩具箱里的玩具太多了.享受周末的人们.

Moral: KISS, there are far too many toys in the toybox these days. Enjoy the weekend people.

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