Re: copy constructor not being called

On 16 Kwi, 19:34, Chris Uzdavinis <cuz...@[EMAIL PROTECTED]
> wrote:
> On Apr 16, 1:22 am, Marcin Swiderski <sfider.b...@[EMAIL PROTECTED]
> wrote:
>
> > On 15 Kwi, 20:33, suresh <suresh.amritap...@[EMAIL PROTECTED]
> wrote:
>
> > >    base b1 = 'x'; //why copy constructor not called here?
>
> > This line calls base(const char) constructor and that is the correct
> > behaviour. This sytanx means that object b1 is initialized with
> > value 'x'. It's equivalent to:
>
> > base b1('x');
>
> It's NOT equivalent, it only seems so most of the time.
>
You learn something every day. I've done a little experiment:

# include <iostream>

class A {
public:
     A(int i) {
         std::cout << "Conversion from integer\n";
     };
private:
     A(const A& rhs) {
         std::cout << "Copy construction\n";
     };
};

int main() {
     A a = 1;
     return 0;
}

This will fail to compile (under gcc 3.4.4) with
the following error:

cnstr.cpp:9: error: `A::A(const A&)' is private

If you delete 'private:' it will compile, but it won't
use the copy constructor.

Furthermore, I've found an example where the use of each
one of the two initialization notations will give a different
output (also under gcc 3.4.4):

# include <iostream>

class B;

class A {
public:
     A() {
         std::cout << "Default construction\n";
     }
     A(const B& b) {
         std::cout << "Conversion from B\n";
     }
     A(const A& rhs) {
         std::cout << "Copy construction\n";
     }
};

class B {
public:
     operator A() {
         std::cout << "Conversion to A\n";
         return A();
     }
};

int main() {
     B b;
     A a = b;
     A c(b);
     return 0;
}

This program will output:

Conversion to A
Default construction
Conversion from B

Cheers
Sfider

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