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From: Andre Woebbeking <[EMAIL PROTECTED]>
To: Debian Bug Tracking System <[EMAIL PROTECTED]>
Subject: g++-4.0: name lookup is broken
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Date: Thu, 09 Dec 2004 22:57:24 +0100
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Package: g++-4.0
Version: 4.0-0pre2
Severity: important
Tags: experimental
Hi,
the following code doesn't compile:
struct A {};
namespace Boo
{
struct B
{
friend struct A;
B(const A&) {};
};
}
int main()
{
A a;
Boo::B b(a);
return 0;
}
It compiles with g++ < 4.0.
Cheers,
André
-- System Information:
Debian Release: 3.1
APT prefers unstable
APT policy: (500, 'unstable'), (1, 'experimental')
Architecture: i386 (i686)
Kernel: Linux 2.6.9-1-k7
Locale: [EMAIL PROTECTED], [EMAIL PROTECTED] (charmap=ISO-8859-15)
Versions of packages g++-4.0 depends on:
ii gcc-4.0 4.0-0pre2 The GNU C compiler
ii gcc-4.0-base 4.0-0pre2 The GNU Compiler Collection (base
ii libc6 2.3.2.ds1-19 GNU C Library: Shared libraries an
ii libstdc++6-4.0-dev 4.0-0pre2 The GNU Standard C++ Library v3 (d
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---------------------------------------
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Date: Tue, 7 Jun 2005 21:32:52 -0400
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Subject: Not a bug
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This isn't a bug. By declaring friend struct A inside of the namespace
Boo, you have created a prototype for it in the *Boo* namespace. This is not
an issue with the
compiler at all. In fact, this is a very strict interpretation of the
code, which is a good thing (TM). To solve the situation, just set it
like this:
struct A {};
namespace Boo
{
struct B
{
friend struct ::A; // Note the :: for global namespace
B(const A&) {};
};
}
int main()
{
A a;
Boo::B b(a);
return 0;
}
By declaring ::A in the friend, it makes it the only prototyped struct A, thus
the constructor will take that A.
Dan Weber
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