c++ - C++11: Template Function Specialization for Integer Types

Question

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?

Answer 1

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)
{
    // ...
}