TITLE: non-template vs template versions of functions

(Newsgroups: comp.lang.c++.moderated, 22 Mar 98)

ZIAUDDIN: Najeem Ziauddin <najeem@nest.stpt.soft.net>

[snip]

> consider the following problem.
> 
> template <class T>
> void
> foo(T)
> {
>         cout<<endl<<"in template"<<endl;
> }
> 
> template <>
> void
> foo<int>(int )
> {
>         cout<<endl<<"in special int"<<endl;
> }
> 
> void
> foo(int)
> {
>         cout<<endl<<"in int"<<endl;
> }
> 
> If i give a statement foo(int()) in the main function , the o/p is "in
> special int". If I put the specialization below the function, then the
> o/p will be "in int".


VANDEVOORDE: David Vandevoorde <daveed@cup.hp.com>

Where is "the main function"?

The fact that you "specialize the function template" is relatively
unimportant. At the point where you call a 'foo' with an 'int'
there are presumably two possibilities:
   (a) only the template version is visible
   (b) both the template and non-template version is visible

If (a), there is no choice and you pick the template
version.

If (b), overload resolution finds that both are equal-rank
candidates, but there is a rule that in that case non-templates
are preferred over templates and hence you pick the non-template.