🚀 go-pugleaf

In article 1bfa@news.doit.wisc.edu, khan@xraylith.wisc.edu (Mumit Khan) writes:
>
>  14.8  Template arguments                                    [temp.arg]
>
>4 Nor is a local type or a type with no linkage name an acceptable  tem-
>  plate-argument.  [Example:
>          void f()
>          {
>              struct S { /* ... */ };                     
>              X<S,p> x3; // error: local type used as template-argument
>          }
>   --end example] 
>
>Why is this necessary? Why isn't the template argument S in X<S,p>
>simply treated as X<f()::S, p> (or something along these lines)?

The problem is "something along those lines". I hope it is clear that the
template parameter would have to be somehow qualified with its scope,
so that some other instantiation with some other type S could be seen
to be a different type.

C++ lacks a mechanism to refer to a local type  outside its scope. Just
qualifying "S" with "f()" would not be enough, because "f()" might be a
static function, and thus there could be other functions "f()" in the program.
If you want to refer to a local type elsewhere, declare the type in a
namespace instead of in a function (thus it is no longer a local type).

So as not to require a method for accessing a local type outside its scope,
templates are not allowed to be instantiated with a local type.
---
Steve Clamage, stephen.clamage@eng.sun.com



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