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Date: Mon, 26 Jan 2004 17:20:00 -0000
From: "joerg dot richter at pdv-fs dot de" <gcc-bugzilla@gcc.gnu.org>
To: gcc-bugs@gcc.gnu.org
Message-ID: <20040126172022.13868.joerg.richter@pdv-fs.de>
Reply-To: gcc-bugzilla@gcc.gnu.org
Subject: [Bug c++/13868] New: Conditional expression operator and implicit char array conversion
X-Bugzilla-Reason: CC
X-SW-Source: 2004-01/txt/msg03291.txt.bz2
List-Id: <gcc-bugs.sourceware.org>

Compile this program with GCC3.3.2:

----

struct Foo
{
    typedef char type[4];
    type mem;

    operator type&()
    {
      return mem;
    }
};

int main()
{
  true ? "ab" : Foo();
  return 0;
}

----

and you get the following output:

a.cc:14: no match for `bool ? const char[5] : Foo' operator
a.cc:14: candidates are: operator?:(bool, char*, char*) <built-in>
a.cc:14:                 operator?:(bool, const char*, const char*) <built-in>


I think this should compile. Using the user defined conversion to char (&) [4] 
and then the array to pointer conversion on both sides. Then the qualification 
conversion and you have the same types char const*.

Removing the need for the array to pointer conversion, by changing "ab" 
to "abc" helps. Or removing the qualification conversion by changing the 
implicit conversion to "operator type const&()" helps also.

-- 
           Summary: Conditional expression operator and implicit char array
                    conversion
           Product: gcc
           Version: 3.3.2
            Status: UNCONFIRMED
          Severity: normal
          Priority: P2
         Component: c++
        AssignedTo: unassigned at gcc dot gnu dot org
        ReportedBy: joerg dot richter at pdv-fs dot de
                CC: gcc-bugs at gcc dot gnu dot org


http://gcc.gnu.org/bugzilla/show_bug.cgi?id=13868