我写第二个形式为：

  template< typename T> 
 auto foo（T a，T b，...） - > decltype（a + = b，void（））
 {
 a + = b; 
} 
$ p> 

  decltype的推导类型（a + = b，如果表达式 a + = b 有效，则void（）void 将只是 void  
 

好吧，即使在第一种形式中，我也会使用尾随返回类型方法。
 
I'm trying to eliminate an overload from an overload set if operator+= is missing.

I know how to check if T+T is legal :
template<typename T,
         typename CheckTplusT = decltype(std::declval<T>() + std::declval<T>())>
void foo(T a, T b, ...)
{
  a = a + b;
}

but this doesn't work for +=
template<typename T,
         typename CheckTplusT = decltype(std::declval<T>() += std::declval<T>())>
void foo(T a, T b, ...)
{
  a += b;
}

Is this fixable by using another expression inside decltype or do I need another SFINAE construct?

The reason I need this eliminated from the overload set is that it ***es with another overload that accepts a functor to be used as an alternative to +=. Compilers are VS2013, gcc4.8

I would write the second form as:
template<typename T>
auto foo(T a, T b, ...) -> decltype( a+=b, void() )
{
  a += b;
}

The deduced type for decltype(a+=b, void()) would be just void if the expression a+=b is valid, else it would result in SFINAE.

Well, even in the first form, I would use the trailing-return type approach.