Bug 50711

Unfortunately, I don't think you can use std::result_of for sfinae purposes, if I understand correctly it's a well know annoyance which you have to overcome by open coding with std::declval.

Eg, if you simply open the <functional> header you will find plenty of decltype( std::declval... for sfinae, which cannot be replaced by a concise std::result_of

The following code works:

struct Tuple
{
  int a, b, c;
};

struct array_get
{
  template <typename T>
  const T&
  operator()(const T& t)
  {
    return t;
  }

  template <typename Array, typename First, typename... Location>
  auto
  operator()(const Array& a, First f, Location... loc)
  //  -> typename std::result_of<array_get(decltype(a[f]), Location...)>::type
    -> decltype(operator()(a[f], loc...))
  {
    return operator()(a[f], loc...);
  }
};

struct Array
{
  int array[5][5][5];

  Array()
  {
    for (int i = 0; i != 5*5*5; ++i)
    {
      reinterpret_cast<int*>(array)[i] = i;
    }
  }

  int get(const Tuple& t)
  {
    return array[t.a][t.b][t.c];
  }

  template <typename... Location>
  //typename std::result_of<array_get(int[5][5][5], Location...)>::type
  auto
  get(Location... loc)
    -> decltype(array_get()(array, loc...))
  {
    return array_get()(array, loc...);
  }
};

int main(int argc, char *argv[])
{
  Array a;
  Tuple t{3,4,1};
  //return a.get(1,2,3);
  return a.get(t);
}

See http://lwg.github.com/issues/lwg-defects.html#1225 and also 1270

result_of cannot be used for SFINAE, it requires that its arguments be a valid expression

(In reply to comment #4)
> result_of cannot be used for SFINAE, it requires that its arguments be a valid
> expression

Should read: it requires that its arguments can form a valid expression

Ok, I guess we can close this.