What does it mean when one says something is SFINAE-friendly?

Question

I can\'t clearly get the grasp of what it means when one mentions that a particular function, struct or ... is SFINAE-friendly.

Would someone please

Answer 1

An entity is termed SFINAE-friendly if it can be used in the context of SFINAE without producing a hard error upon substitution failure. I assume you already know what SFINAE is, as that is a whole other question in itself.

In the context of C++ standardization, the term SFINAE-friendly has so far been applied to std::result_of and std::common_type. Take the following example:

template 
void foo(T x, typename std::common_type::type y) {}

void foo(std::string x, std::string y) {}

int main()
{
    foo(std::string("hello"), std::string("world"));
}

Without SFINAE-friendly common_type, this would fail to compile, because std::common_type::type would produce a hard error during template argument substitution. With the introduction of SFINAE-friendly common_type (N3843) this example becomes well-formed, because std::common_type::type produces a substitution failure so that overload is excluded from the viable set.

Here's a similar example with result_of:

template 
auto bar(T f) -> typename std::result_of::type { return f(); }

void bar(int n) {}

int main()
{
    bar(42);
}

Without SFINAE-friendly result_of, this would fail to compile, because std::result_of::type would produce a hard error during template argument substitution. With the introduction of SFINAE-friendly result_of (N3462) this example becomes well-formed, because std::result_of::type produces a substitution failure so that overload is excluded from the viable set.