C++ - What is the purpose of function template specialization? When to use it?

Question

Learning C++, came upon function templates. The chapter mentioned template specialization.

1. template <> void foo(int);

Answer 1

Multiple overloads on the same name do similar things. Specializations do the exact same thing, but on different types. Overloads have the same name, but may be defined in different scopes. A template is declared in only one scope, and the location of a specialization declaration is insignificant (although it must be at the scope of the enclosing namespace).

For example, if you extend std::swap to support your type, you must do so by specialization, because the function is named std::swap, not simply swap, and the functions in would be quite right to specifically call it as ::std::swap( a, b );. Likewise for any name that might be aliased across namespaces: calling a function may get "harder" once you qualify the name.

The scoping issue is confused further by argument-dependent lookup. Often an overload may be found because it is defined in proximity to the type of one of its arguments. (For example, as a static member function.) This is completely different from how the template specialization would be found, which is by simply looking up the template name, and then looking up the explicit specialization once the template has been chosen as the target of the call.

The rules of ADL are the most confusing part of the standard, so I prefer explicit specialization on the priciple of avoiding reliance on it.