假设您有以下两个课程:
class A
{
public:
int a;
int b;
}
class B
{
public:
int a;
int b;
}
class C
{
public:
float a1;
float b1;
}
enum class Side { A, B };
我想要一个模板函数,它需要一个side
and a T
,并且取决于T
,返回对“的引用T.a
" or "T.b
“如果班级有成员T::a
,或引用“T.a1
" or "T.b1
“如果班级有成员T::a1
.
我的出发点是:
template<typename T>
auto &GetBySide(const Side &side, const T &twoSided)
{
return side == Side::A?twoSided.a:twoSided.b;
}
template<typename T>
auto &GetBySide(const Side &side, const T &twoSided)
{
return side == Side::A?twoSided.a1:twoSided.b1;
}
问题是如何让编译器跳过第一个模板,如果成员a
不存在。
所以我实现了下面@Jarod42给出的解决方案,但它在VS 2015中给出了错误,因为VS区分模板的能力存在错误。这是一个解决方法:
template<typename T>
auto GetBySide(const Side &side, const T& twoSided)
-> decltype((twoSided.a))
{
return side == Side::A ? twoSided.a : twoSided.b;
}
// Using comma operator to trick compiler so it doesn't think that this is the same as above
template<typename T>
auto GetBySide(const Side &side, const T &twoSided)
-> decltype((0, twoSided.a1))
{
return side == Side::A ? twoSided.a1 : twoSided.b1;
}
// See comment above
template<typename T>
auto GetBySide(const Side &side, const T &twoSided)
-> decltype((0, 0, twoSided.a2))
{
return side == Side::A ? twoSided.a2 : twoSided.b2;
}
另一种方法是使用逗号运算符和代表每个“概念”的特殊结构