Re: assignment operator using a constructor

On Apr 18, 2:00 pm, "Jeff Baker" <algort...@[EMAIL PROTECTED]
> wrote:
> How does an assignment operator work? Following the code that I give
below,
> I notice it isn't clear how the base operator works. I am not sure. It
seems
> to me that there is a copy constructor formed that is used after the '='
is
> call. Cout <<
> "assigment operator ", prints first the at the return the copy
constuctor is
> called. Is there a copy constructor created before the print  statement
is
> executed or is it during the return a.s? Or what is going on?
>
> Here is the code:
>
> #ifndef H_H
> #define H_H
>   # include <iostream>
> using namespace std;
> class base
> {
> private:
>     char s;
> public:
>     base( ){}
>     base(const char a ):s(a) {cout << "construction with arg " << int(a)
<<
> endl;}
>     base(const base & a){cout <<"copy constructor " << int(a.s) <<
endl;}
>     base  operator=(const base & a){cout << "assignment operator  " <<
a.s <<
> endl; return a.s;}
>     ~base( ) {cout << "destruction" << endl;}};
>
> #endif
> #include "h.h"
> int main()
> {
>   base b3;
>   base b1 = 'x'; // is equivalent to  base b1('x');
>   base b2 = b1; //copy constructor is called Is seen as Base b2(b1)
>   b3 = b1; // assignment operator called and copy constructor is called
>
>   return 0;
>
> }

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base  operator=(const base & a)
{
     cout << "assignment operator  " << a.s << endl;
     return a.s;
}

Your problem is caused because of the object you are returning.
You return the character a.s but the return type is a base: So you are
getting a call to a constructor to make the return object. You are
also returning the object so you will also get a call to copy
construct the object out of the function.

What I expect you meant was:
base&  operator=(const base & a)
{
     cout << "assignment operator  " << a.s << endl;
     return *this;
}

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