懒惰评价

Question

我该如何懒惰评估std :: conditional中的第二个参数？

#include "stdafx.h" 
#include <type_traits> 

struct Null{}; 
struct _1{enum {one = true,two = false};}; 
struct _2{enum {two = true, one = false};}; 

template<class T> 
struct is_nulltype 
{ 
    enum {value = false}; 
}; 

template<> 
struct is_nulltype<Null> 
{ 
    enum {value = true}; 
}; 

template<class T> 
struct X : std::conditional<is_nulltype<T>::value,Null,typename std::conditional<T::one,_1,_2>::type>::type 
{ 
}; 

int _tmain(int argc, _TCHAR* argv[]) 
{ 
X<Null> x;//won't compile no Null::one but I don't need that member in Null at all 
    return 0; 
} 

Answer 1

这种情况的常用的方法是有std::conditional 2元功能之间进行选择：两个::type小号std::conditional后现在

template <typename T> 
struct false_case { 
    typedef typename std::conditional<T::one,_1,_2>::type type; 
}; 

struct always_null {typedef Null type;}; 

template<class T> 
struct X : 
    std::conditional<is_nulltype<T>::value, 
        always_null, 
        false_case<T> 
        >::type::type { ... }; 

注意。