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Message-ID:	<405A2B70.80800@phekda.gotadsl.co.uk>
Date:	Thu, 18 Mar 2004 23:06:24 +0000
From:	Richard Dawe <ng AT phekda DOT gotadsl DOT co DOT uk>
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Newsgroups:	comp.os.msdos.djgpp
Subject:	Re: iterator depricated warning
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To:	djgpp AT delorie DOT com
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Hello.

DanG2015 wrote:
> I'm sure there's a simple answer to this, but I can't find it.
> 
> The code:========================
> 
> #include <deque>
> 
> template <class T>
> class foo {
> 
> public:
>     std::deque<T>   foo_deque;
> 
>     void foo_method( void ) {
>         std::deque<T>::iterator i = foo_deque.begin();
>     }
> };
> 
> int main( void ) {
> }
> 
> the compile=========================
> 
> C:> gpp -Wall test.cpp
> test.cpp: In member function `void foo<T>::foo_method()':
> test.cpp:10: warning: `typename std::deque<T, std::allocator<_CharT>
> 
>>::iterator' is implicitly a typename
> 
> test.cpp:10: warning: implicit typename is deprecated, please see the
> documentation for details
[snip]

I've seen this before. At a guess:

std::deque<> takes a type name, e.g.: int or SomeClass. You need to get 
the type name from the template parameter. E.g.:

   std::deque<typename T>   foo_deque;

and:

   std::deque<typename T>::iterator i = foo_deque.begin();

I guess that implicit typenames are not supported, to remove ambiguity. 
Consider the clash between a template parameter T and a type called T. 
I'm just guessing - I don't have my copy of Stroustrop here.

Bye, Rich =]

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