Template Function Specialization for Integer Types

c++, templates

Solution

Use SFINAE

// For all types except integral types:
template<typename T>
typename std::enable_if<!std::is_integral<T>::value>::type f(T t)
{
    // ...
}

// For integral types only:
template<typename T>
typename std::enable_if<std::is_integral<T>::value>::type f(T t)
{
    // ...
}

Note that you will have to include the full `std::enable_if` return value even for the declaration.

C++17 update:

// For all types except integral types:
template<typename T>
std::enable_if_t<!std::is_integral_v<T>> f(T t)
{
    // ...
}

// For integral types only:
template<typename T>
std::enable_if_t<std::is_integral_v<T>> f(T t)
{
    // ...
}

Problem

Suppose I have a template function: ``` template<typename T> void f(T t) { ... } ``` and I want to write a specialization for all primitive integer types. What is the best way to do this? What I mean is: ``` template<typename I where is_integral<I>::value is true> void f(I i) { ... } ``` and the compiler selects the second version for integer types, and the first version for everything else?

Original source