My overlaoded assignment operator doesn't work

[heng]

I define my own assignment operator, but it doesn't work as I imagined.

#include<iostream>
#include <string>
using namespace std;

class A{
int x;
public:
int y;
int z;
A& operator=(const A& t)
{
z=t.y;
y=t.z;
return *this;
}
};

int main()
{
A aclass;
aclass.y=9;
aclass.z=11;
A bclass=aclass;
cout<<aclass.z<<endl<<bclass.y;
string str;
cin>>str;
return 0;
}

 

[Rolf Magnus]

I define my own assignment operator, but it doesn't work as I imagined.

That's unfortunate. What did you imagine, and what does it do instead?

#include<iostream>
#include <string>
using namespace std;

class A{
int x;
public:
int y;
int z;
A& operator=(const A& t)
{
z=t.y;
y=t.z;
return *this;
}
};

int main()
{
A aclass;
aclass.y=9;
aclass.z=11;
A bclass=aclass;
cout<<aclass.z<<endl<<bclass.y;
string str;
cin>>str;
return 0;
}

This program never calls operator=. Note that

A bclass=aclass;

is not assigment, but initialization.

 

[heng]

This program never calls operator=. Note that


is not assigment, but initialization.

Thanks. How may I call operator=? Could you please show me an example?

And I am kind of confused about assignment and initialization. Is the
following belonging to assignment?

A bclass;
bclass=aclass;

 

[Rolf Magnus]

Thanks. How may I call operator=? Could you please show me an example?

And I am kind of confused about assignment and initialization.

It's rather simple:

Type Object = OtherObject; <- initialization
Object = OtherObject; <- assignment

In other words: if you define a new object, it's initialization, if you have
an already existing object on the left side, it's assignment.

Is the following belonging to assignment?

A bclass;
bclass=aclass;

Yes.

 

[Kai-Uwe Bux]

I define my own assignment operator, but it doesn't work as I imagined.

#include<iostream>
#include <string>
using namespace std;

class A{
int x;
public:
int y;
int z;
A& operator=(const A& t)
{
z=t.y;
y=t.z;

I would expect:

z = t.z;
y = t.y;

Are you sure you want that?

return *this;
}
};

int main()
{
A aclass;
aclass.y=9;
aclass.z=11;
A bclass=aclass;

This calls the copy constructor.

cout<<aclass.z<<endl<<bclass.y;
string str;
cin>>str;
return 0;
}

Best

Kai-Uwe Bux

 

[heng]

Thanks.

I would expect:

z = t.z;
y = t.y;

Are you sure you want that?


This calls the copy constructor.



Best

Kai-Uwe Bux

 

[Jacek Dziedzic]

I define my own assignment operator, but it doesn't work as I imagined.

#include<iostream>
#include <string>
using namespace std;

class A{
int x;
public:
int y;
int z;
A& operator=(const A& t)
{
z=t.y;
y=t.z;
return *this;
}
};

int main()
{
A aclass;
aclass.y=9;
aclass.z=11;
A bclass=aclass;
cout<<aclass.z<<endl<<bclass.y;
string str;
cin>>str;
return 0;
}

This invokes the copy constructor, not the assignment
operator. Why not define that one too?

HTH,
- J.

 

[TvN]

Hi,

I define my own assignment operator, but it doesn't work as I imagined.

[skiped]

People already said why it was not called, my reply about your
operator Why do you check for self assigment? What about such
situation:

A a;
....
a = a;

So I think you should always check for this:

A& operator=(const A& t)
{
if (this != &t)
{
....
}
return *this;
}

Regards,
Volodymyr!

 

[red floyd]

Hi,

I define my own assignment operator, but it doesn't work as I imagined.

[skiped]

People already said why it was not called, my reply about your
operator Why do you check for self assigment? What about such
situation:

A a;
...
a = a;

So I think you should always check for this:

A& operator=(const A& t)
{
if (this != &t)
{
....
}
return *this;
}

See Sutter's "Exceptional C++" for a discussion of self-assignment
checks. It should only be necessary as an optimization, if you use the
construct-and-swap metaphor.