Approaches to function SFINAE in C++ [duplicate]

Question

I am using function SFINAE heavily in a project and am not sure if there are any differences between the following two approaches (other than style):

#include <cstdlib>
#include <type_traits>
#include <iostream>

template <class T, class = std::enable_if_t<std::is_same_v<T, int>>>
void foo()
{
    std::cout << "method 1" << std::endl;
}

template <class T, std::enable_if_t<std::is_same_v<T, double>>* = 0>
void foo()
{
    std::cout << "method 2" << std::endl;
}

int main()
{
    foo<int>();
    foo<double>();

    std::cout << "Done...";
    std::getchar();

    return EXIT_SUCCESS;
}

The program output is as expected:

method 1
method 2
Done...

I have seen method 2 used more often in stackoverflow, but I prefer method 1.

Are there any circumstances when these two approaches differ?

Lamination

How do you run this program? It won't compile for me. – Pinnace

@alter igel it will need a C++17 compiler. I used MSVC 2019 to test this example but I mainly work with Clang. – Lamination

Related: why-should-i-avoid-stdenable-if-in-function-signatures and C++20 introduces also new ways with concept :-) – Erichericha

@Erichericha Concepts are one of most needed things for me from C++20. – Incommodious

Answer 1

I have seen method 2 used more often in stackoverflow, but I prefer method 1.

Suggestion: prefer method 2.

Both methods work with single functions. The problem arises when you have more than a function, with the same signature, and you want enable only one function of the set.

Suppose that you want enable foo(), version 1, when bar<T>() (pretend it's a constexpr function) is true, and foo(), version 2, when bar<T>() is false.

With

template <typename T, typename = std::enable_if_t<true == bar<T>()>>
void foo () // version 1
 { }

template <typename T, typename = std::enable_if_t<false == bar<T>()>>
void foo () // version 2
 { }

you get a compilation error because you have an ambiguity: two foo() functions with the same signature (a default template parameter doesn't change the signature).

But the following solution

template <typename T, std::enable_if_t<true == bar<T>(), bool> = true>
void foo () // version 1
 { }

template <typename T, std::enable_if_t<false == bar<T>(), bool> = true>
void foo () // version 2
 { }

works, because SFINAE modify the signature of the functions.

Unrelated observation: there is also a third method: enable/disable the return type (except for class/struct constructors, obviously)

template <typename T>
std::enable_if_t<true == bar<T>()> foo () // version 1
 { }

template <typename T>
std::enable_if_t<false == bar<T>()> foo () // version 2
 { }

As method 2, method 3 is compatible with selection of alternative functions with same signature.

Weal

Thanks for the great explanation, I will prefer methods 2 & 3 from now on :-) – Lamination

"a default template parameter doesn't change the signature" - how is this different in your second variant, which also uses default template parameters? – Neurologist

@Neurologist - Non simple to say... I suppose the other answer explain this better... If SFINAE enable/disable the default template argument, the foo() function remain available when you call it with an explicit second template parameter (the foo<double, double>(); call). And if remain available, there is an ambiguity with the other version. With method 2, SFINAE enable/disable the second argument, not the default parameter. So you can't call it explicating the parameter because there is a substitution failure that doesn't permit a second parameter. So the version is unavailable, so no ambiguity – Weal

Method 3 has the additional advantage of generally not leaking into the symbol-name. The variant auto foo() -> std::enable_if_t<...> is often useful to avoid hiding the function-signature and to allow using the function-arguments. – Cultus

@max66: so the key point is that substitution failure in a template parameter default is not an error if the parameter is supplied and no default is needed? – Neurologist

@Neurologist - yes... seems to me that you caught the point: a substitution failure for a template parameter default disables only the calls of the function that don't express the parameter but maintain available the function itself. Method 2 disable that function itself. – Weal

Answer 2

In addition to max66's answer, another reason to prefer method 2 is that with method 1, you can (accidentally) pass an explicit type parameter as the second template argument and defeat the SFINAE mechanism completely. This could happen as a typo, copy/paste error, or as an oversight in a larger template mechanism.

#include <cstdlib>
#include <type_traits>
#include <iostream>

// NOTE: foo should only accept T=int
template <class T, class = std::enable_if_t<std::is_same_v<T, int>>>
void foo(){
    std::cout << "method 1" << std::endl;
}

int main(){

    // works fine
    foo<int>();

    // ERROR: subsitution failure, as expected
    // foo<double>();

    // Oops! also works, even though T != int :(
    foo<double, double>();

    return 0;
}

Live demo here

Pinnace