Question

Let’s assume we have a template function “foo”:

template<class T>
void foo(T arg)
{ ... }

I can make specialization for some particular type, e.g.

template<>
void foo(int arg)
{ ... }

If I wanted to use the same specialization for all builtin numeric types (int, float, double etc.) I would write those lines many times. I know that body can be thrown out to another function and just call of this is to be made in every specialization’s body, however it would be nicer if i could avoid writting this “void foo(…” for every type. Is there any possibility to tell the compiler that I want to use this specialization for all this types?

Answer 1

You could use std::numeric_limits to see whether a type is a numeric type (is_specialized is true for all float and integer fundamental types).

// small utility
template<bool> struct bool2type { };

// numeric
template<typename T>
void fooImpl(T arg, bool2type<true>) {

}

// not numeric
template<typename T>
void fooImpl(T arg, bool2type<false>) {

}

template<class T>
void foo(T arg)
{ fooImpl(arg, bool2type<std::numeric_limits<T>::is_specialized>()); }