c ++函子和函数模板 [英] c++ functor and function templates

问题描述

考虑这个简单和无意义的代码。

consider this simple and pointless code.

#include <iostream>

struct A {
    template<int N>
    void test() {
    	std::cout << N << std::endl;
    }
};

int main() {
    A a;
    a.test<1>();
}

这是一个非常简单的函数模板示例。然而，如果我想用重载的运算符（）替换 A :: test 以使其成为函子？

It is a very simple example of a function template. What if however, I wanted to replace A::test with an overloaded operator() to make it a functor?

#include <iostream>

struct A {
    template<int N>
    void operator()() {
    	std::cout << N << std::endl;
    }
};

int main() {
    A a;
    a<1>(); // <-- error, how do I do this?
}

当然，如果 $ c>参数取决于模板，编译器可能推断出模板。但是我只是无法找出合适的语法来使用无参数函数指定模板参数。

Certainly if the operator() took parameters which were dependent on the template, the compiler could possibly deduce the template. But I just can't figure out the proper syntax to specify template parameters with a parameterless functor.

有正确的方法吗？ >

Is there a proper way to do this?

显然，这段代码可以工作，因为它绕过了functor语法：

Obviously, this code would work since it bypasses the functor syntax:

a.operator()<1>();

但是这样做会失去它作为函子的目的:-P。

but that kinda defeats the purpose of it being a functor :-P.

推荐答案

除了：

 a.operator()<1>();

语法。如果你打算更改代码，将模板参数移动到类将工作，或使用（boost | tr1）::绑定，使（boost | tr1）::函数对象。

syntax. If you're open to changing the code, moving the template parameter to the class would work, or using a (boost|tr1)::bind to make a (boost|tr1)::function object.

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