Question

The STL functors are implemented like this:

template<class T>
struct less{
  bool operator()(T const& lhs, T const& rhs){
    return lhs < rhs;
  }
};

This makes us mention the (possibly long) type everytime we create such a functor. Why are they not implemented like shown below? Any reasons?

struct less{
  template<class T>
  bool operator()(T const& lhs, T const& rhs){
    return lhs < rhs;
  }
};

That would make them usable without any mentioning of (possibly long) types.

Answer 1

It would also make it impossible to specialize them for user defined types.

They are supposed to be a customization point.

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To summarize the discussions in the comments:

Although it is technically possible to do like Xeo suggests, the language standard doesn’t allow it.

It is very hard to write a working class template if users are allowed to specialize individual functions of the template. In some cases it might however be a good idea to specialize the whole class for a user defined type.

Therefore the C++98 standard writes (17.4.3.1):

It is undefined for a C++ program to add declarations or definitions to namespace std or namespaces within namespace std unless otherwise specified. A program may add template specializations for any standard library template to namespace std.

As it isn’t “otherwise specified” that Xeo’s code is allowed, we are to understand that it is not. Perhaps not totally obvious! Or that “template specializations” only apply to classes.

The new C++11 standard has had this part expanded, and spells it out in more detail (17.6.4.2):

The behavior of a C++ program is undefined if it adds declarations or definitions to namespace std or to a namespace within namespace std unless otherwise specified. A program may add a template specialization for any standard library template to namespace std only if the declaration depends on a user-defined type and the specialization meets the standard library requirements for the original template and is not explicitly
prohibited.

The behavior of a C++ program is undefined if it declares

— an explicit specialization of any member function of a standard library class template, or
— an explicit specialization of any member function template of a standard library class or class template, or
— an explicit or partial specialization of any member class template of a standard library class or class template.

A program may explicitly instantiate a template defined in the standard library only if the declaration depends on the name of a user-defined type and the instantiation meets the standard library requirements for the original template.