Re: Operator Cast () Reference?

From:
"Francesco S. Carta" <entuland@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Wed, 9 Jun 2010 19:56:16 -0700 (PDT)
Message-ID:
<2bc5c73e-d0a7-4f4b-80fd-5070526160f6@u7g2000yqm.googlegroups.com>
"Francesco S. Carta" <entul...@gmail.com> wrote:

Twixer Xev <twixer_...@hotmail.com> wrote:

I should clarify. Your assignments never invoke the cast operator. Put a
trace in there and you'll see what I mean. Assignment is covered by the
ctor's, which happen to be identicle for each class.


Ah, I see what you mean (silly me). In fact, if I declare the
constructors as explicit both assignments halt the compiler. Thanks
for the pointer :-)


In addition, if I got it straight:

-------
#include <iostream>

using namespace std;

class IntByRef {
    public:
        explicit IntByRef(int i) : datum(i) {};
        operator int&() {
            return datum;
        }
    private:
        int datum;
};

class IntByVal {
    public:
        explicit IntByVal(int i) : datum(i) {};
        operator int() {
            return datum;
        }
    private:
        int datum;
};

int main()
{
    int one = 1;
    int two = 2;

    IntByRef intbyref(one);

    IntByVal intbyval(two);

    cout << intbyref << endl; // prints 1

    int& rint = intbyref;
    rint = 42;
    cout << intbyref << endl; // prints 42

    // int& rint2 = intbyval; // chokes the compiler (*)

    const int& crint = intbyval; // fine (**)
    const int& crint2 = 78; // fine (**)

    intbyval = IntByVal(10000);

    cout << crint << endl; // prints 2
    cout << crint2 << endl; // prints 78
    cout << intbyval << endl; // prints 10000

    return 0;
}
-------

(*) The compiler complains because it is not possible to create a non-
const reference to something that isn't modifiable.

(**) Fine? Seems so. Compiles and gives the expected result ;-)

--
FSC
http://userscripts.org/scripts/show/59948

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