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) {     // ... } 

Answer 2

I would use overload resolution. That spares you from having to use the gross SFINAE hack. Unfortunately there are many areas where you can't avoid it, but this fortunately isn't one of those.

template<typename T> void f(T t) {   f(t, std::is_integral<T>()); }  template<typename T> void f(T t, std::true_type) {    // ... }  template<typename T> void f(T t, std::false_type) {    // ... }