根据成员的存在选择模板功能

假设你有这两个类:

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 };

我希望有一个模板函数,该函数side和T,并根据T返回的引用," T.a"或" T.b"如果类有一个成员T::a,或参考" T.a1"或" T.b1"如果类有一个成员T::a1.

我的出发点是:

template
auto &GetBySide(const Side &side, const T &twoSided) 
{ 
  return side == Side::A?twoSided.a:twoSided.b; 
}

template
auto &GetBySide(const Side &side, const T &twoSided) 
{ 
  return side == Side::A?twoSided.a1:twoSided.b1; 
}

问题是如果成员a不存在,如何让编译器跳过第一个模板.

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能够区分模板.这是一个解决方法:

template
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
auto GetBySide(const Side &side, const T &twoSided) 
-> decltype((0, twoSided.a1))
{ 
  return side == Side::A ? twoSided.a1 : twoSided.b1; 
}

// See comment above
template
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表示每个"概念"的特殊结构

有了SFINAE.

template
auto GetBySide(const Side &side, const T& twoSided) 
-> decltype((twoSided.a))
{ 
  return side == Side::A ? twoSided.a : twoSided.b; 
}

template
auto GetBySide(const Side &side, const T &twoSided) 
-> decltype((twoSided.a1))
{ 
  return side == Side::A ? twoSided.a1 : twoSided.b1; 
}

演示