SFINAE with C++14 return type deduction

问题:

Thanks to C++14, we'll soon be able to curtail verbose trailing return types; such as the generic min example from David Abrahams 2011 post:

template <typename T, typename U>
auto min(T x, U y)
    -> typename std::remove_reference< decltype(x < y ? x : y) >::type
{ return x < y ? x : y; }

Under C++14 the return type can be omitted, and min can be written as:

template <typename T, typename U>
auto min(T x, U y)
{ return x < y ? x : y; }

This is a simple example, however return type deduction is very useful for generic code, and can avoid much replication. My question is, for functions such as this, how do we integrate SFINAE techniques? For example, how can I use std::enable_if to restrict our min function to return types which are integral?

回答1:

You can't SFINAE the function using the return type if you're using return type deduction. This is mentioned in the proposal

Since the return type is deduced by instantiating the template, if the instantiation is ill-formed, this causes an error rather than a substitution failure.

You can, however, use an extra, unused template parameter to perform SFINAE.

template <class T, class U>
auto min1(T x, U y)
{ return x < y ? x : y; }

template <class T, class U,
          class...,
          class = std::enable_if_t<std::is_integral<T>::value &&
                                   std::is_integral<U>::value>>
auto min2(T x, U y)
{
    return x < y ? x : y; 
}

struct foo {};

min1(foo{}, foo{}); // error - invalid operands to <
min1(10, 20);

min2(foo{}, foo{}); // error - no matching function min2
min2(10, 20);

Live demo