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: 
Use [SFINAE](http://en.wikipedia.org/wiki/Substitution_failure_is_not_an_error)

```
// 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)
{
    // ...
}

```
Here is C++20 solution

```cpp
template<std::integral T>
void f(T v) {
   ...
}

```

I think It's actually overloaded function but works the same.