How to SFINAE to include a member function returning `auto`
In a template metaprogram, you can use SFINAE on return type to select a specific template member function, i.e.
template<int N> struct A {
int sum() const noexcept
{ return _sum<N-1>(); }
private:
int _data[N];
template<int I> typename std::enable_if< I,int>::type _sum() const noexcept
{ return _sum<I-1>() + _data[I]; }
template<int I> typename std::enable_if<!I,int>::type _sum() const noexcept
{ return _data[I]; }
};
However, this will not work if the function _sum()
in question ( in the above example) has an auto-detected return type, such _func()
as in this example
template<int N> class A
{
/* ... */
private:
// how to make SFINAE work for _func() ?
template<int I, typename BinaryOp, typename UnaryFunc>
auto _func(BinaryOp op, UnaryFunc f) const noexcept -> decltype(f(_data[0]))
{ return op(_func<I-1>(op,f),f(_data[I])); }
};
What else can you do to get SFINAE here?
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1 answer
Following on from David Rodriguez-Dribe, the following code worked as intended:
template<int N> class A
{
int min_abs() const noexcept
{
return _func<N-1>([](int x, int y)->int { return std::min(x,y); },
[](int x)->int { return std::abs(x); });
}
private:
int _data[N];
template<int I, typename BinaryOp, typename UnaryFunc>
auto _func(BinaryOp op, UnaryFunc f) const noexcept
-> typename std::enable_if< I>,decltype(f(_data[0]))>::type
{ return op(_func<I-1>(op,f),f(_data[I])); }
template<int I, typename BinaryOp, typename UnaryFunc>
auto _func(BinaryOp op, UnaryFunc f) const noexcept
-> typename std::enable_if<!I>,decltype(f(_data[0]))>::type
{ return f(_data[I]); }
};
strictly speaking, we also need to make sure that the binary operator op
returns the correct type. For simplicity and brevity of the answer, I leave this for the reader to understand ...
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