Re: assignment operator using a constructor

Using this to start. base& and *this is the correct form.
   base  operator=(const base & a){cout << "assignment operator  " << a.s
<< 
endl; return a.s;}

I am trying to understand how each return type works. I am introducing an 
int return type on next line.
int operator=(const base& a){cout << "asignment operator " << a.s << endl;

return a.s};

What is created if base is change to int? Like base operator =(/...../){} 
copy constructor created. As base& operator=(/.../){}; there is not copy 
constructor
As int return type there no extra constructor and seem equivalent to
 base&  operator=(const base & a)
{
    cout << "assignment operator  " << a.s << endl;
    return *this;
The reference  base& doen't create a copy constructor and there is only a 
temp object for the return.

Which is more efficient the int operator as written or the base type as 
written?
I understand that it might be preferred to use one over the other
determined 
how a problem is handled.

Jeff
>
> 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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