模板检查类成员函数的存在? [英] Templated check for the existence of a class member function?
问题描述
是否可以编写一个模板,根据类上是否定义了某个成员函数来改变行为?
Is it possible to write a template that changes behavior depending on if a certain member function is defined on a class?
这是我想写的一个简单例子:
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";
}
所以,如果class T
定义了toString()
,那么它就使用它;否则,它不会.我不知道该怎么做的神奇部分是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
是的,通过 SFINAE,您可以检查给定的类是否确实提供了某种方法.这是工作代码:
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;
struct two { char x[2]; };
template <typename C> static one test( decltype(&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 对其进行了测试.我不知道它是否可以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.
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