How Does SFINAE Behavior Differ Between Return Types and Template Parameters in C Function Overloads?

SFINAE Behavior Variance in Return Types vs. Template Parameters

In C , the Substitute Failure Is Not an Error (SFINAE) idiom allows conditional function overloads based on template metaprogramming and type traits. However, there are instances where SFINAE behaves differently depending on its placement in the function template declaration.

Consider the following code:

template<typename T,
         typename = typename std::enable_if<
                        std::is_integral<T>::value>::type>
void foo(T);  // Incorrect placement, triggers error

template<typename T,
         typename = typename std::enable_if<
                        std::is_floating_point<T>::value>::type>
void foo(T);  // Incorrect placement, triggers error

template<typename T>
void foo(T)    // Correct placement
    -> typename std::enable_if<
            std::is_integral<T>::value>::type;

template<typename T>
void foo(T)    // Correct placement
    -> typename std::enable_if<
            std::is_floating_point<T>::value>::type;

In the first set of overloads (incorrect placement), placing SFINAE in the template parameters rather than the return type results in a compilation error. This is because default template arguments (in this case, the ::type suffix) are not considered in determining overload equivalence. Consequently, the compiler treats the functions as duplicate declarations with the same signature.

However, in the second set of overloads (correct placement), SFINAE is applied to the return type, which involves template metaprogramming. This placement ensures that the expression referring to template parameters is part of the function signature. As a result, the compiler recognizes the overloads as distinct entities, allowing SFINAE to function properly.

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