Assignment operator

[al]

string s = "original string";

s = "new string";
s.assign("anoth er new string");

Is there any difference using above two way to assign a new string to s?
class Derived : public Base
{
};

Base b;
Derived d;
list<Base> mylist;
mylist.push_bac k(d);

Is assignment operator called? If true, whose operator is called, Base's or
Derived's?

b = d;

Whose assignment operator called, Base's or Derived's?

Thank you very much!

[Cy Edmunds]

"al" <al***@168.ne t> wrote in message
news:hX******** *************** @bgtnsc04-news.ops.worldn et.att.net...

string s = "original string";

s = "new string";
s.assign("anoth er new string");

Is there any difference using above two way to assign a new string to s?
No.


class Derived : public Base
{
};

Base b;
Derived d;
list<Base> mylist;
mylist.push_bac k(d);

Is assignment operator called?
No, d should be copied into a new node using Base's copy constructor.
If true, whose operator is called, Base's or
Derived's?

b = d;

Whose assignment operator called, Base's or Derived's?
Base's.

Thank you very much!

You're welcome. Hope I got everything right! :)

--
Cy
http://home.rochester.rr.com/cyhome/

[Dan Cernat]

"al" <al***@168.ne t> wrote in message
news:hX******** *************** @bgtnsc04-news.ops.worldn et.att.net...

string s = "original string";

s = "new string";
s.assign("anoth er new string");

Is there any difference using above two way to assign a new string to s?
the result is the same: s gets the new value. how it is done is
implementation dependant.


class Derived : public Base
{
};

Base b;
Derived d;
list<Base> mylist;
mylist.push_bac k(d);

Is assignment operator called? If true, whose operator is called, Base's or Derived's?

b = d;

Whose assignment operator called, Base's or Derived's?
yes, the assignment operator is called. Since you store Base objects in your
list, the d object will be trimmed down into a Base object and the
Base::operator= will be called. thje list will store only teh Base portion
of the d object.

to store various objects in a container, use pointers to their base (they
have to have the same base)

list <Base *> mySmartList;

Thank you very much!

HTH
Dan

[Dan Cernat]

I read Cy's answer and I realised that I gave one answer for two questions.
See below

"Dan Cernat" <ce****@dan.com > wrote in message
news:vu******** ****@corp.super news.com...

"al" <al***@168.ne t> wrote in message
news:hX******** *************** @bgtnsc04-news.ops.worldn et.att.net...

string s = "original string";

s = "new string";
s.assign("anoth er new string");

Is there any difference using above two way to assign a new string to s?
the result is the same: s gets the new value. how it is done is
implementation dependant.

class Derived : public Base
{
};

Base b;
Derived d;
list<Base> mylist;
mylist.push_bac k(d);

Is assignment operator called? If true, whose operator is called, Base's or

Derived's?

Nope, it is the copy constructor called (as Cy said)
b = d;

Whose assignment operator called, Base's or Derived's?

yes, the assignment operator is called. Since you store Base objects in

your list, the d object will be trimmed down into a Base object and the
Base::operator= will be called. thje list will store only teh Base portion
of the d object. The paragraph above should be (disregard the mumbling about list):
yes the assignment operator is called. the d object will be trimmed down
into a Base object and the Base::operator= will be called.

to store various objects in a container, use pointers to their base (they
have to have the same base)

list <Base *> mySmartList;
this is correct

Thank you very much!

HTH
Dan

Dan

[Jeff Schwab]

al wrote:

string s = "original string";

s = "new string";
s.assign("anoth er new string");

Is there any difference using above two way to assign a new string to s?

Grep your header. Here's what mine says:

basic_string&
operator=(const basic_string& __str) { return this->assign(__str ); }