Assignment operator

[al]

string s = "original string";

s = "new string";
s.assign("another 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_back(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]

string s = "original string";

s = "new string";
s.assign("another 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_back(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!

 

[Dan Cernat]

string s = "original string";

s = "new string";
s.assign("another 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_back(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:perator= 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)

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

the result is the same: s gets the new value. how it is done is
implementation dependant.
Base'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:perator= 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:perator= 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

HTH
Dan

Dan

 

[Jeff Schwab]

string s = "original string";

s = "new string";
s.assign("another 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); }