Re: C++ problem

[Shachar Shemesh]

Oleg Goldshmidt wrote:
Well, while it was making the round trip to the list and back, I also
made a trip to the bookshelf to check myself. Item 15 of Scott
Meyers's "Effective C++" confirms what I wrote: temps are const, and
for the reasons I covered. It explicitly says that with a declaration
like
    C& C::operator=(C& rhs) { ... }
client code won't compile.
     

It appears, however, that client code can fail to compile, and yet temps 
will not be const.

Then you haven't understood my "dual existence" statement. Try out the
following program:
   

#include <iostream>
using namespace std;
class a {
public:
   void function() const {
       cout<<"Const version called"<<endl;
   }
   void function() {
       cout<<"Non-const version called"<<endl;
   }
   static void func2( const a & ) {
       cout<<"Called with const reference"<<endl;
   }
   static void func2( a & ) {
       cout<<"Called with non-const reference"<<endl;
   }
};
int main()
{
   a().function();
   a::func2(a());
   return 0;
}
     

A temp is a non-const, but it cannot be bound to a non-const
reference. I don't like it, but that seems to be the case.
   

I don't understand this statement. In "a().function()" your temp is
not const, according to your wishes. In "a::func2(a())" the
compiler-generated temp is const, as appropriate.
Then please explain what makes the first and the second different.
If you don't believe
me, comment out the declaration of "void func2(const a&)" (you are
trying to call "void func2( a & )" if I understand your drift
correctly) and try to compile.
 

While it is indeed nice of you to point out to me the very exact same 
fact I'm trying to point out to you and say "told you so". In other 
words, I really lost you.

There is no difference between the first a() and the second a(). Both 
are temporary variables of type a generated by an empty constructor. 
However, we can see from the sample program that this variable can be 
used for a non-const method, thus plainly contradicting that temporary 
variables are const. Then again, we see from the second line that it 
cannot be bound to a non-const reference. Thus, you get my claim that 
they have a duel semantics.

This is in complete agreement with what I wrote before (and what is
still quoted above).
 

No. You said temps are const. We can see from the first line of the 
sample program above that they are clearly not (at least, not always).

         Shachar
--
Shachar Shemesh
Lingnu Open Source Consulting ltd.
http://www.lingnu.com/
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