我有以下模板化对象:
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;
};
我该如何使用std::enable_if
根据需要启用或禁用构造函数?
e.g:
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
This似乎有效,但我不确定这是最佳方法
因此,只需将具有默认值的新模板参数添加到构造函数即可启用 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,
typename std::enable_if<!std::is_same<T1, 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, typename T2 = type_2>
result( type_1 f, 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);
}
另请阅读:使用enable_if选择类构造函数
本文内容由网友自发贡献,版权归原作者所有,本站不承担相应法律责任。如您发现有涉嫌抄袭侵权的内容,请联系:hwhale#tublm.com(使用前将#替换为@)