我有以下模板对象：
template< typename type_1,typename type_2 > struct result
{
// I want to enable these two constructors only if type_1 != type_2
result( type_1 f ) : foo{f} {}
result( type_2 b ) : bar{b} {}
// I want to enable this constructor only if type_1 == type_2
result( type_1 f,type_2 b ) : foo{f},bar{b} {}
// Other member functions removed.
type_1 foo;
type_2 bar;
};
如何根据需要使用std :: enable_if启用或禁用构造函数？
例如：
这个只有前两个构造函数：
result<string,int> // type_1 != type_2
这个只有第三个构造函数：
result<int,int> // type_1 == type_2
解决方法
This似乎有效,但我不确定它是最佳方式
因此,只需将具有默认值的新模板参数添加到构造函数以启用SFINAE
#include <type_traits>
template< typename type_1,typename type_2 >
struct result
{
// I want to enable these two constructors only if type_1 != type_2
template<typename T1 = type_1,typename T2 = type_2>
result( type_1 f,typename std::enable_if<!std::is_same<T1,T2>::value>::type * = nullptr )
: foo{f} {}
template<typename T1 = type_1,typename T2 = type_2>
result( type_2 b,T2>::value,int >::type * = nullptr )
: bar{b} {} /* ^^^ need this to avoid duplicated signature error with above one*/
// I want to enable this constructor only if type_1 == type_2
template<typename T1 = type_1,type_2 b,typename std::enable_if<std::is_same<T1,T2>::value>::type * = nullptr )
: foo{f},bar{b} {}
type_1 foo;
type_2 bar;
};
int main()
{
result<int,double> r(1);
result<int,double> r2(1.0);
result<int,int> r3(1,2);
// disbaled
//result<int,double> r4(1,2.0);
//result<int,int> r5(1);
}
