Re: Is this a -Weffc++ bug in gcc?

From:

red floyd <no.spam.here@its.invalid>

Newsgroups:

comp.lang.c++

Date:

Thu, 16 Aug 2012 15:42:53 -0700

Message-ID:

<k0jt1c$73p$2@dont-email.me>

On 8/16/2012 3:41 PM, red floyd wrote:

On 8/16/2012 2:51 PM, Scott Lurndal wrote:

Victor Bazarov <v.bazarov@comcast.invalid> writes:

On 8/16/2012 4:53 PM, DeMarcus wrote:

If I compile the following with gcc 4.7.1 and -Weffc++ then I get a
compiler warning on the operator% but not the others. Is this a bug?

struct A
{
     A& operator%( int i ) { return *this; }
     A& operator++() { return *this; }
     A& fnc( int i ) { return *this; }
};

int main()
{
     return 0;
}

And for those of us who don't have [any intention to use] gcc, what
compiler warning do you get? Just out of curiosity, of course...

$ g++ -Weffc++ -o /tmp/a.o -c /tmp/a.cc
/tmp/a.cc:3:25: warning: 'A& A::operator%(int)' should return by value
[-Weffc++]

Because the semantics of a binary arithmetic operator such as "%"
indicate that the return should be an rvalue. Scott Meyers discusses
this in "Effective C++" (which is what the -Weffc++ looks for).

Consider: Does the following make sense?

int x, y;
(x % y) = 3;

If, however, the following does make sense:

A a;
(a + 3) = 7;

Oops. Typed too fast.

A a, b;
(a % 3) = b;

Then feel free to ignore the warning.