if-else取决于T是否为完整类型 [英] if-else depends on whether T is a complete type

问题描述

如何检查特定类型.cpp中特定类型是否为完整类型?

How to check whether a certain type is a complete type in a certain .cpp?

template<class T>class Test{
    //some fields
    void(*functor)(T*) =[](T*){}; 
    //^ will be written by some .cpp that can access T as complete-type 
    T* t=nullptr;
    void fComplete(){    
        delete t;     //faster
        /** ^ some code that use complete type*/    
    }
    void fForward(){
        functor(t);   //slower
        /** ^ some code that forward declaration is enough*/   
    }
    void f(){  
        /*if(T is complete type){    
            fComplete();
        }else fForward();*/
    }
};

演示

当我想过早地优化自定义智能指针中的删除功能时，它将很有用.

It will be useful when I want to prematurely optimize a delete function in my custom smart pointer.

任何人都可以确认这是不可能的吗?
我不是要解决方法(但我不介意)-这个问题只是我的好奇心.

Can anyone confirm that it is impossible?
I am not asking for a workaround (but I don't mind) - this question is just my curiosity.

推荐答案

有效

#include <iostream>
#include <type_traits>

using namespace std;

class Incomplete;
class Complete {};

template <typename IncompleteType, typename = std::enable_if_t<true>>
struct DetermineComplete {
    static constexpr const bool value = false;
};

template <typename IncompleteType>
struct DetermineComplete<
        IncompleteType,
        std::enable_if_t<sizeof(IncompleteType) == sizeof(IncompleteType)>> {
    static constexpr const bool value = true;
};

int main() {
    cout << DetermineComplete<Complete>::value << endl;
    cout << DetermineComplete<Incomplete>::value << endl;
    return 0;
}

注意:我喜欢使用std::enable_if_t来达到与void_t相同的效果，直到可用为止，而不是自己到处编写它的实现.

Note I like to use std::enable_if_t for the same effect as void_t until that is available instead of writing its implementation myself everywhere.

注意也请查看有关ODR的其他答案.它们提出了您在使用此功能之前应考虑的有效观点.

Note Do take a look at the other answer as well about ODR. They bring up a valid point that you should consider before using this.

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