是否可以编写一个模板来检查函数的存在？ [英] Is it possible to write a template to check for a function's existence?

问题描述

是否可以写一个模板来改变行为，取决于是否在类上定义了某个成员函数？

Is it possible to write a template that changes behavior depending on if a certain member function is defined on a class?

这里是一个简单的例子，写：

Here's a simple example of what I would want to write:

template<class T>
std::string optionalToString(T* obj)
{
    if (FUNCTION_EXISTS(T->toString))
        return obj->toString();
    else
        return "toString not defined";
}

因此，如果 class T 有 toString（）定义，然后它使用它;否则，它不会。我不知道该怎么做的魔法部分是FUNCTION_EXISTS部分。

So, if class T has toString() defined, then it uses it; otherwise, it doesn't. The magical part that I don't know how to do is the "FUNCTION_EXISTS" part.

推荐答案

检查一个给定的类是否提供了一个特定的方法。这里是工作代码：

Yes, with SFINAE you can check if a given class does provide a certain method. Here's the working code:

#include <iostream>

struct Hello
{
    int helloworld() { return 0; }
};

struct Generic {};    

// SFINAE test
template <typename T>
class has_helloworld
{
    typedef char one;
    typedef long two;

    template <typename C> static one test( typeof(&C::helloworld) ) ;
    template <typename C> static two test(...);    

public:
    enum { value = sizeof(test<T>(0)) == sizeof(char) };
};

int main(int argc, char *argv[])
{
    std::cout << has_helloworld<Hello>::value << std::endl;
    std::cout << has_helloworld<Generic>::value << std::endl;
    return 0;
}

我刚刚用Linux和gcc 4.1 / 4.3测试过。我不知道它是否可移植到运行不同编译器的其他平台。

I've just tested it with Linux and gcc 4.1/4.3. I don't know if it's portable to other platforms running different compilers.

这篇关于是否可以编写一个模板来检查函数的存在？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！