Re: c++/10146: [3.4 regression] [new parser] template function lookup failure(s)

Re: c++/10146: [3.4 regression] [new parser] template function lookup failure(s)

[Richard Guenther]

The following reply was made to PR c++/10146; it has been noted by GNATS.

From: Richard Guenther <rguenth@tat.physik.uni-tuebingen.de>
To: Giovanni Bajo <giovannibajo@libero.it>
Cc: gcc-gnats@gcc.gnu.org, <gcc-bugs@gcc.gnu.org>, <nobody@gcc.gnu.org>,
   <gcc-prs@gcc.gnu.org>
Subject: Re: c++/10146: [3.4 regression] [new parser] template function lookup
 failure(s)
Date: Wed, 19 Mar 2003 13:20:42 +0100 (CET)

 On Wed, 19 Mar 2003, Giovanni Bajo wrote:
 
 >
 > http://gcc.gnu.org/cgi-bin/gnatsweb.pl?cmd=view%20audit-trail&database=gcc&p
 > r=10146
 >
 > To sum it up:
 >
 > >Foo<int>().template foo<U>(u); // does not work
 > >Foo<int>().template bar<U>(u); // does not work
 >
 > These should compile.
 >
 > >Foo<int>().foo(u);             // does work ??
 > >Foo<int>::foo(u);              // does work ??
 > >Foo<int>().bar(u);             // does work ??
 >
 > Yes, because the template parameter of the template member function is
 > deduced from the call. What's wrong with them?
 
 I think foo() and bar() needs to be qualified with the template keyword
 due to two-stage namelookup. But I may be wrong (dont have a standard
 to look at).
 
 Richard.
 
 --
 Richard Guenther <richard.guenther@uni-tuebingen.de>
 WWW: http://www.tat.physik.uni-tuebingen.de/~rguenth/
 

Re: c++/10146: [3.4 regression] [new parser] template function lookup failure(s)

[bangerth]

Synopsis: [3.4 regression] [new parser] template function lookup failure(s)

State-Changed-From-To: open->analyzed
State-Changed-By: bangerth
State-Changed-When: Thu Mar 20 02:04:59 2003
State-Changed-Why:
    Giovanni did this already

http://gcc.gnu.org/cgi-bin/gnatsweb.pl?cmd=view%20audit-trail&database=gcc&pr=10146

Re: c++/10146: [3.4 regression] [new parser] template function lookup failure(s)

[Richard Guenther]

The following reply was made to PR c++/10146; it has been noted by GNATS.

From: Richard Guenther <rguenth@tat.physik.uni-tuebingen.de>
To: Gabriel Dos Reis <gdr@integrable-solutions.net>
Cc: Giovanni Bajo <giovannibajo@libero.it>, <gcc-gnats@gcc.gnu.org>,
   <gcc-bugs@gcc.gnu.org>
Subject: Re: c++/10146: [3.4 regression] [new parser] template function lookup
 failure(s)
Date: Wed, 19 Mar 2003 13:47:40 +0100 (CET)

 On 19 Mar 2003, Gabriel Dos Reis wrote:
 
 > Richard Guenther <rguenth@tat.physik.uni-tuebingen.de> writes:
 >
 > | On 19 Mar 2003, Gabriel Dos Reis wrote:
 > |
 > | > Richard Guenther <rguenth@tat.physik.uni-tuebingen.de> writes:
 > | >
 > | > | On Wed, 19 Mar 2003, Giovanni Bajo wrote:
 > | > |
 > | > | >
 > | > | > http://gcc.gnu.org/cgi-bin/gnatsweb.pl?cmd=view%20audit-trail&database=gcc&p
 > | > | > r=10146
 > | > | >
 > | > | > To sum it up:
 > | > | >
 > | > | > >Foo<int>().template foo<U>(u); // does not work
 > | > | > >Foo<int>().template bar<U>(u); // does not work
 > | > | >
 > | > | > These should compile.
 > | > | >
 > | > | > >Foo<int>().foo(u);             // does work ??
 > | > | > >Foo<int>::foo(u);              // does work ??
 > | > | > >Foo<int>().bar(u);             // does work ??
 > | > | >
 > | > | > Yes, because the template parameter of the template member function is
 > | > | > deduced from the call. What's wrong with them?
 > | > |
 > | > | I think foo() and bar() needs to be qualified with the template keyword
 > | > | due to two-stage namelookup. But I may be wrong (dont have a standard
 > | > | to look at).
 > | >
 > | > Two-phase name lookup has nothing to do here.  The Foo<int> part is
 > | > -not- dependent, therefore there need not be any extra "template"
 > | > qualifier.
 > |
 > | Ah, ok - only for Foo<T>:: it would be dependent, yes?
 >
 > Yes.  (Any dependent expression would do also).
 
 If I exchange Foo<int> with Foo<T> only
   Foo<T>().foo<U>(u);
 and
   Foo<T>().bar<U>(u);
 do not work (as expected), but f.i. Foo<T>::foo<U>(u); (while
 Foo<int>::foo<U>(u) is rejected) is ok?
 
 Well, I'll leave it now up to Mark to check&fix all relevant cases
 after 3.3 is out.
 
 Richard.
 
 --
 Richard Guenther <richard.guenther@uni-tuebingen.de>
 WWW: http://www.tat.physik.uni-tuebingen.de/~rguenth/
 

Re: c++/10146: [3.4 regression] [new parser] template function lookup failure(s)

[Giovanni Bajo]

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The following reply was made to PR c++/10146; it has been noted by GNATS.

From: "Giovanni Bajo" <giovannibajo@libero.it>
To: "Richard Guenther" <rguenth@tat.physik.uni-tuebingen.de>
Cc: <gcc-gnats@gcc.gnu.org>,
	<gcc-bugs@gcc.gnu.org>,
	<nobody@gcc.gnu.org>,
	<gcc-prs@gcc.gnu.org>
Subject: Re: c++/10146: [3.4 regression] [new parser] template function lookup failure(s)
Date: Wed, 19 Mar 2003 13:53:38 +0100

 No, because Foo<int>::foo (and the others) is not a dependent name. §14.2p4
 e p5 explain that the keyword "template" to qualify member templates must be
 used if the postfix expression (obj.member in short) is dependent, and must
 never be used to qualify members which are not templates.
 So, in those three lines basically the template keyword is optional, but not
 required.
 
 Giovanni Bajo
 

Re: c++/10146: [3.4 regression] [new parser] template function lookup failure(s)

[Gabriel Dos Reis]

The following reply was made to PR c++/10146; it has been noted by GNATS.

From: Gabriel Dos Reis <gdr@integrable-solutions.net>
To: Richard Guenther <rguenth@tat.physik.uni-tuebingen.de>
Cc: Giovanni Bajo <giovannibajo@libero.it>, <gcc-gnats@gcc.gnu.org>,
   <gcc-bugs@gcc.gnu.org>
Subject: Re: c++/10146: [3.4 regression] [new parser] template function lookup failure(s)
Date: 19 Mar 2003 13:40:44 +0100

 Richard Guenther <rguenth@tat.physik.uni-tuebingen.de> writes:
 
 | On 19 Mar 2003, Gabriel Dos Reis wrote:
 | 
 | > Richard Guenther <rguenth@tat.physik.uni-tuebingen.de> writes:
 | >
 | > | On Wed, 19 Mar 2003, Giovanni Bajo wrote:
 | > |
 | > | >
 | > | > http://gcc.gnu.org/cgi-bin/gnatsweb.pl?cmd=view%20audit-trail&database=gcc&p
 | > | > r=10146
 | > | >
 | > | > To sum it up:
 | > | >
 | > | > >Foo<int>().template foo<U>(u); // does not work
 | > | > >Foo<int>().template bar<U>(u); // does not work
 | > | >
 | > | > These should compile.
 | > | >
 | > | > >Foo<int>().foo(u);             // does work ??
 | > | > >Foo<int>::foo(u);              // does work ??
 | > | > >Foo<int>().bar(u);             // does work ??
 | > | >
 | > | > Yes, because the template parameter of the template member function is
 | > | > deduced from the call. What's wrong with them?
 | > |
 | > | I think foo() and bar() needs to be qualified with the template keyword
 | > | due to two-stage namelookup. But I may be wrong (dont have a standard
 | > | to look at).
 | >
 | > Two-phase name lookup has nothing to do here.  The Foo<int> part is
 | > -not- dependent, therefore there need not be any extra "template"
 | > qualifier.
 | 
 | Ah, ok - only for Foo<T>:: it would be dependent, yes?
 
 Yes.  (Any dependent expression would do also).
 
 -- Gaby

Re: c++/10146: [3.4 regression] [new parser] template function lookup failure(s)

[Gabriel Dos Reis]

The following reply was made to PR c++/10146; it has been noted by GNATS.

From: Gabriel Dos Reis <gdr@integrable-solutions.net>
To: Richard Guenther <rguenth@tat.physik.uni-tuebingen.de>
Cc: Giovanni Bajo <giovannibajo@libero.it>, gcc-gnats@gcc.gnu.org,
   <gcc-bugs@gcc.gnu.org>
Subject: Re: c++/10146: [3.4 regression] [new parser] template function lookup failure(s)
Date: 19 Mar 2003 13:26:14 +0100

 Richard Guenther <rguenth@tat.physik.uni-tuebingen.de> writes:
 
 | On Wed, 19 Mar 2003, Giovanni Bajo wrote:
 | 
 | >
 | > http://gcc.gnu.org/cgi-bin/gnatsweb.pl?cmd=view%20audit-trail&database=gcc&p
 | > r=10146
 | >
 | > To sum it up:
 | >
 | > >Foo<int>().template foo<U>(u); // does not work
 | > >Foo<int>().template bar<U>(u); // does not work
 | >
 | > These should compile.
 | >
 | > >Foo<int>().foo(u);             // does work ??
 | > >Foo<int>::foo(u);              // does work ??
 | > >Foo<int>().bar(u);             // does work ??
 | >
 | > Yes, because the template parameter of the template member function is
 | > deduced from the call. What's wrong with them?
 | 
 | I think foo() and bar() needs to be qualified with the template keyword
 | due to two-stage namelookup. But I may be wrong (dont have a standard
 | to look at).
 
 Two-phase name lookup has nothing to do here.  The Foo<int> part is
 -not- dependent, therefore there need not be any extra "template"
 qualifier. 
 
 -- Gaby

Re: c++/10146: [3.4 regression] [new parser] template function lookup failure(s)

[Richard Guenther]

The following reply was made to PR c++/10146; it has been noted by GNATS.

From: Richard Guenther <rguenth@tat.physik.uni-tuebingen.de>
To: Gabriel Dos Reis <gdr@integrable-solutions.net>
Cc: Giovanni Bajo <giovannibajo@libero.it>, <gcc-gnats@gcc.gnu.org>,
   <gcc-bugs@gcc.gnu.org>
Subject: Re: c++/10146: [3.4 regression] [new parser] template function lookup
 failure(s)
Date: Wed, 19 Mar 2003 13:32:12 +0100 (CET)

 On 19 Mar 2003, Gabriel Dos Reis wrote:
 
 > Richard Guenther <rguenth@tat.physik.uni-tuebingen.de> writes:
 >
 > | On Wed, 19 Mar 2003, Giovanni Bajo wrote:
 > |
 > | >
 > | > http://gcc.gnu.org/cgi-bin/gnatsweb.pl?cmd=view%20audit-trail&database=gcc&p
 > | > r=10146
 > | >
 > | > To sum it up:
 > | >
 > | > >Foo<int>().template foo<U>(u); // does not work
 > | > >Foo<int>().template bar<U>(u); // does not work
 > | >
 > | > These should compile.
 > | >
 > | > >Foo<int>().foo(u);             // does work ??
 > | > >Foo<int>::foo(u);              // does work ??
 > | > >Foo<int>().bar(u);             // does work ??
 > | >
 > | > Yes, because the template parameter of the template member function is
 > | > deduced from the call. What's wrong with them?
 > |
 > | I think foo() and bar() needs to be qualified with the template keyword
 > | due to two-stage namelookup. But I may be wrong (dont have a standard
 > | to look at).
 >
 > Two-phase name lookup has nothing to do here.  The Foo<int> part is
 > -not- dependent, therefore there need not be any extra "template"
 > qualifier.
 
 Ah, ok - only for Foo<T>:: it would be dependent, yes? So gcc should
 reject or at least warn about the extra "template" here? (gcc3.2 bails
 out with a parse error for Foo<int>::template foo(u);)
 
 Richard.
 
 --
 Richard Guenther <richard.guenther@uni-tuebingen.de>
 WWW: http://www.tat.physik.uni-tuebingen.de/~rguenth/
 

Re: c++/10146: [3.4 regression] [new parser] template function lookup failure(s)

[Giovanni Bajo]

The following reply was made to PR c++/10146; it has been noted by GNATS.

From: "Giovanni Bajo" <giovannibajo@libero.it>
To: <gcc-gnats@gcc.gnu.org>,
	<rguenth@tat.physik.uni-tuebingen.de>,
	<gcc-bugs@gcc.gnu.org>,
	<nobody@gcc.gnu.org>,
	<gcc-prs@gcc.gnu.org>
Cc:  
Subject: Re: c++/10146: [3.4 regression] [new parser] template function lookup failure(s)
Date: Wed, 19 Mar 2003 13:18:09 +0100

 http://gcc.gnu.org/cgi-bin/gnatsweb.pl?cmd=view%20audit-trail&database=gcc&p
 r=10146
 
 To sum it up:
 
 >Foo<int>().template foo<U>(u); // does not work
 >Foo<int>().template bar<U>(u); // does not work
 
 These should compile.
 
 >Foo<int>().foo(u);             // does work ??
 >Foo<int>::foo(u);              // does work ??
 >Foo<int>().bar(u);             // does work ??
 
 Yes, because the template parameter of the template member function is
 deduced from the call. What's wrong with them?
 
 Note: I did not verify since I cannot compile GCC 3.4.
 
 Giovanni Bajo
 

c++/10146: [3.4 regression] [new parser] template function lookup failure(s)

[rguenth]

>Number:         10146
>Category:       c++
>Synopsis:       [3.4 regression] [new parser] template function lookup failure(s)
>Confidential:   no
>Severity:       critical
>Priority:       medium
>Responsible:    unassigned
>State:          open
>Class:          rejects-legal
>Submitter-Id:   net
>Arrival-Date:   Wed Mar 19 08:56:00 UTC 2003
>Closed-Date:
>Last-Modified:
>Originator:     Richard Guenther
>Release:        g++-3.4 (GCC) 3.4 20030313 (experimental)
>Organization:
>Environment:

>Description:
In the following testcase the marked lines fail/work even if they should/should not (the former being the problem for me). One should carefuly check the standard if any of the working one should be rejected.

The error message from compiling the testcase is

bellatrix:~/src/tests$ g++-3.4 -c temp.cpp 
temp.cpp: In member function `void Bar<T>::operator()(const U&) [with U = int, 
   T = char]':
temp.cpp:32:   instantiated from here
temp.cpp:14: error: no matching function for call to `Foo<int>::foo(const int&)
   '
temp.cpp:15: error: no matching function for call to `Foo<int>::foo(const int&)
   '
temp.cpp:16: error: no matching function for call to `Foo<int>::foo(const int&)
   '
temp.cpp:32:   instantiated from here
temp.cpp:18: error: no matching function for call to `Foo<int>::bar(const int&)
   '
temp.cpp:19: error: no matching function for call to `Foo<int>::bar(const int&)
   '


Testcase:

template <class T>
struct Foo {
        template <class U>
        static void foo(const U&);
        template <class U>
        void bar(const U&);
};

template <class T>
struct Bar {
        template <class U>
        void operator()(const U&u)
        {
                Foo<int>().foo<U>(u);          // does not work (ok)
                Foo<int>::foo<U>(u);           // does not work (ok)
                Foo<int>().template foo<U>(u); // does not work
                Foo<int>::template foo<U>(u);  // does work
                Foo<int>().bar<U>(u);          // does not work (ok)
                Foo<int>().template bar<U>(u); // does not work
                Foo<int>().foo(u);             // does work ??
                Foo<int>::foo(u);              // does work ??
                Foo<int>().template foo(u);    // does work
                Foo<int>::template foo(u);     // does work
                Foo<int>().bar(u);             // does work ??
                Foo<int>().template bar(u);    // does work
        }
};

void foo()
{
        int i;
        Bar<char>()(i);
}
>How-To-Repeat:

>Fix:

>Release-Note:
>Audit-Trail:
>Unformatted: