具有std :: is_base_of的派生类的C ++模板函数 [英] C++ template function for derived class with std::is_base_of

问题描述

如果给定类型的函数是从另一个函数派生的，而对于所有其他情况，则创建其他函数，我会遇到问题.我的代码:

I've got problem with creating function that for given type, if it's derived from other one do something and for all other cases do something other. My code:

class BaseClass {};
class DerivedClass : public BaseClass {};

template <typename T>
void Function(typename std::enable_if<std::is_base_of<BaseClass, T>::value, T>::type && arg) {
    std::cout << "Proper";
}

template <typename T>
void Function(T && arg) {
    std::cout << "Improper";
}

void test() {
    Function(DerivedClass{});
}

对于类DeriviedClass和其他基于BaseClass的类，我想调用函数 couting Proper，但是它是 couts Improper.有什么建议吗?

For class DeriviedClass and other based on BaseClass I'd like to call function couting Proper, but it couts Improper. Any suggestions?

推荐答案

正如问题注释中所提到的那样，SFINAE表达式无法像您那样使用.
它应该是这样的:

As mentioned in the comments to the question, SFINAE expressions won't work the way you did it.
It should be instead something like this:

template <typename T>
typename std::enable_if<std::is_base_of<BaseClass, T>::value>::type
Function(T && arg) {
    std::cout << "Proper" << std::endl;
}

template <typename T>
typename std::enable_if<not std::is_base_of<BaseClass, T>::value>::type
Function(T && arg) {
    std::cout << "Improper" << std::endl;
}

SFINAE表达式将启用或禁用Function，具体取决于BaseClass是T的基础.在这两种情况下，返回类型均为void，因为如果未定义，则它是std::enable_it的默认类型.
在 coliru 上查看.

SFINAE expressions will enable or disable Function depending on the fact that BaseClass is base of T. Return type is void in both cases, for it's the default type for std::enable_it if you don't define it.
See it on coliru.

存在其他有效的替代方法，其中一些已在其他答案中提及.

Other valid alternatives exist and some of them have been mentioned in other answers.

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