Re: template iterator

From:

bob@coolgroups.com

Newsgroups:

microsoft.public.vc.language

Date:

Wed, 30 Jan 2008 20:44:07 -0800 (PST)

Message-ID:

<222bc7a2-3184-4801-9731-62ef72a64949@k2g2000hse.googlegroups.com>

On Jan 30, 4:50 pm, "Doug Harrison [MVP]" <d...@mvps.org> wrote:

On Wed, 30 Jan 2008 16:26:12 -0800 (PST), b...@coolgroups.com wrote:

Why doesn't this compile in VS 2005?

#include "stdafx.h"
#include <vector>

using namespace std;

template < typename ty >

class MyTemplateClass
{
    vector<ty> a;
    vector<ty>::iterator b;
};

int _tmain(int argc, _TCHAR* argv[])
{

    MyTemplateClass<int> test;
    return 0;
}

The "iterator" in vector<ty>::iterator depends on a template parameter and=

is thus said to be a "dependent type". You have to tell the compiler that
"iterator" really is a type as opposed to a non-type member of vector<ty>.=

You do this with the typename keyword:

typename vector<ty>::iterator b;

This is required because vector<ty> could be specialized at the point of
instantiation of MyTemplateClass<ty>, such that iterator could be a type o=

non-type, irrespective of its definition in the primary vector template. B=

default, the compiler assumes it's a non-type, and using the typename
keyword overrides this default.

--
Doug Harrison
Visual C++ MVP- Hide quoted text -

- Show quoted text -

Ok, it compiles now. Out of curiosity, why does the original code

compile in both VC++ 6.0 and Dev-C++ (but not VS 2005)? Is it just a

difference in default assumptions?

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