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我正在尝试使用 mem_fun_ref 将对对象的成员函数的引用发送到另一个函数，但我得到了error C2064: term does not evaluate to a function taking 0 arguments.

我没有在示例中反映这一点，但我需要将 mem_fun_ref_t 发送到一个虚函数，这就是为什么我不只是让 Flip 成为一个接受简单函数对象的函数模板。

#include <iostream>
#include <string>
#include <functional>

class Coin
{
public:
    Coin() {}
    std::string Flip ()
    {
        srand(23);
        int side = rand() % 2 + 1;

        std::string result = "";
        if (side == 1)
            result = "heads.";
        else
            result = "tails.";
        return result;
    }
};



std::string Flip(std::mem_fun_ref_t<std::string, Coin> flip)
{
    return flip();
}



int main()
{
    std::cout << "Flipping a coin..." << std::endl;
    std::string output = Flip(std::mem_fun_ref<std::string, Coin>(&Coin::Flip));
    std::cout << "The coin came up " << output << std::endl;
    return 0;
}
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1 回答 1

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您应该阅读静态成员函数以及成员函数指针。有三种方法可以解决您的问题。

首先是制作Coin::Flip一个静态成员函数：

#include <string>
#include <iostream>

typedef std::string (*Flipper)(); // Function pointer typedef

class Coin
{
public:
    Coin() {}

    // Static member function. A pointer to a static member function can be
    // held in a regular function pointer.
    static std::string Flip ()
    {
        srand(23);
        int side = rand() % 2 + 1;
        return (side == 1) ? "heads." : "tails.";
    }
};

std::string Flip(Flipper flipper)
{
    return flipper();
}

int main()
{
    std::cout << "Flipping a coin..." << std::endl;
    std::string output = Flip(&Coin::Flip);
    std::cout << "The coin came up " << output << std::endl;
    return 0;
}

如果Coin::Flip需要是一个非静态成员函数，你可以传递一个Coin实例Flip，连同成员函数指针：

#include <functional>
#include <string>
#include <iostream>

class Coin
{
public:
    Coin() {}

    // Non-static member function.
    std::string Flip ()
    {
        srand(23);
        int side = rand() % 2 + 1;
        return (side == 1) ? "heads." : "tails.";
    }
};

typedef std::mem_fun_ref_t<std::string, Coin> Flipper;

// We need the Coin instance as well as the member function pointer.
std::string Flip(Coin& coin, Flipper flipper)
{
    // Invoke the flipper member function on the coin instance
    return flipper(coin);
}

int main()
{
    // Since we're using a non-static member function, we need an instance
    // of Coin.
    Coin coin;
    std::cout << "Flipping a coin..." << std::endl;
    std::string output = Flip(coin, mem_fun_ref(&Coin::Flip));
    std::cout << "The coin came up " << output << std::endl;
    return 0;
}

最后，如果Flipper函子可以是来自任何类型对象（不仅是 Coin）的成员函数，并且您不希望Flip自由函数成为模板，那么您将需要std::function并且std::bind这是最近 C++11 的一部分标准。std::function是一个通用的多态函数包装器，适用于任何类型的可调用目标：自由函数、成员函数、函数对象等。如果你不能使用 C++11，Boost 库有等价物boost::function和boost::bind.

#include <functional>
#include <string>
#include <iostream>

class Coin
{
public:
    Coin() {}

    // Non-static member function.
    std::string Flip ()
    {
        srand(23);
        int side = rand() % 2 + 1;
        return (side == 1) ? "heads." : "tails.";
    }

    // Static member function.
    static std::string StaticFlip()
    {
        srand(23);
        int side = rand() % 2 + 1;
        return (side == 1) ? "heads." : "tails.";
    }
};

// Flipper is a generic function object wrapper that works with free functions,
// function objects, static member functions, and non-static member functions.
typedef std::function<std::string ()> Flipper;

std::string Flip(Flipper flipper)
{
    return flipper();
}

int main()
{
    // Example with non-static member function
    Coin coin;

    // Bind a Coin instance along with a Coin::Flip member function pointer.
    Flipper flipper1 = std::bind(&Coin::Flip, &coin);

    std::cout << "Flipping a coin..." << std::endl;
    std::string output = Flip(flipper1);
    std::cout << "The coin came up " << output << std::endl;

    // Example with static member function
    Flipper flipper2 = &Coin::StaticFlip;
    std::cout << "Flipping a coin..." << std::endl;
    output = Flip(flipper2);
    std::cout << "The coin came up " << output << std::endl;

    return 0;
}
于 2012-07-10T16:17:14.403 回答