overloading ->

[ranjit mario noronha]

Hi,

I am trying to overload the ->operator as follows :


class foo {
public:
int a;
int operator->() ;

};
int foo::operator->() {

cout<<"Hullo"<< endl;
return 1l;
}
void main() {
foo *f=new foo();
cout<<f->a;
}

However the value of a is printed (0) and the operator function is not
called. Does anybody know what mistake I'm making ?

Thanks,
__

Ranjit

_______________ _______________ _______________ _______________ _______________ _____

Ranjit Noronha
Graduate Research Associate
Dept. of Computer and Information Sciences
The Ohio State University
Phone: (614)477-9900
E-mail: noronha@cis.ohi o-state.edu

_______________ _______________ _______________ _______________ _______________ _____

[Bernd Jochims]

Re: overloading -&gt;


"ranjit mario noronha" <noronha@xi.cis .ohio-state.edu> wrote in message
news:Pine.GSO.4 .40.03090915253 60.271-100000@xi.cis.o hio-state.edu...[color=blue]
>
> Hi,
>
> I am trying to overload the ->operator as follows :
>
>
> class foo {
> public:
> int a;
> int operator->() ;
>
> };
> int foo::operator->() {
>
> cout<<"Hullo"<< endl;
> return 1l;
> }
> void main() {
> foo *f=new foo();
> cout<<f->a;
> }
>
> However the value of a is printed (0) and the operator function is not
> called. Does anybody know what mistake I'm making ?
>
> Thanks,
> __
>
> Ranjit
>
>[/color]
_______________ _______________ _______________ _______________ _______________ _
____[color=blue]
>
> Ranjit Noronha
> Graduate Research Associate
> Dept. of Computer and Information Sciences
> The Ohio State University
> Phone: (614)477-9900
> E-mail: noronha@cis.ohi o-state.edu
>
>[/color]
_______________ _______________ _______________ _______________ _______________ _
____[color=blue]
>
>///[/color]
try changing
foo *f=new foo();
to
foo* f= new foo;

also, void main should be int main.
Bernie

[Kevin Goodsell]

Re: overloading -&gt;

ranjit mario noronha wrote:
[color=blue]
> Hi,
>
> I am trying to overload the ->operator as follows :
>
>
> class foo {
> public:
> int a;
> int operator->() ;[/color]

operator-> can't return an int. It has to return something that '->' can
be applied to (usually a pointer).
[color=blue]
>
> };
> int foo::operator->() {
>
> cout<<"Hullo"<< endl;
> return 1l;
> }
> void main() {[/color]

In C++, main returns int. void is not and never has been acceptable.
[color=blue]
> foo *f=new foo();
> cout<<f->a;[/color]

f->a is basically equivalent to this:

(f.operator->())->a

This is why operator-> cannot return an int, and must return something
that -> can be applied to. Read about operator-> in your C++ book. It's
not like other operators.

Also, since you did not initialize the foo you allocated, nearly any use
of it (other than giving it a value) invokes undefined behavior. You
can't "use" a variable that has never been assigned a value.

And you forgot to delete f.
[color=blue]
> }
>
> However the value of a is printed (0) and the operator function is not
> called. Does anybody know what mistake I'm making ?
>
> Thanks,
> __
>
> Ranjit
>
> _______________ _______________ _______________ _______________ _______________ _____
>
> Ranjit Noronha
> Graduate Research Associate
> Dept. of Computer and Information Sciences
> The Ohio State University
> Phone: (614)477-9900
> E-mail: noronha@cis.ohi o-state.edu
>
> _______________ _______________ _______________ _______________ _______________ _____
>
>[/color]

Think you could trim that monster down a bit? A sig with more than about
4 or 5 lines is usually considered excessive. Wouldn't hurt if you used
a correct sig marker, either ("-- " is used to indicate the start of a
signature).

-Kevin
--
My email address is valid, but changes periodically.
To contact me please use the address from a recent posting.

[Ron Natalie]

Re: overloading -&gt;

ranjit mario noronha wrote:
[color=blue]
> I am trying to overload the ->operator as follows :[/color]

You've got a lot of problems here.

[color=blue]
> int operator->() ;[/color]

This isn't legal. Operator-> must return something that is legal to apply
another operator -> to. An int does not meet these requirements.

Essentially, when operator-> is overloaded f->a gets interpretted
as (f.operator->())->a
[color=blue]
> void main() {[/color]

main must return int.
[color=blue]
> foo *f=new foo();
> cout<<f->a;[/color]

This doesn't even attempt to call foo::operator->(). You invoked the operator
on a foo*. You can't overload on pointers (only classes and enums). If you
had written:
foo f;
cout << f->a;
things might have been different (of course, with your operator-> it still won't
compile).

[Bernd Jochims]

Re: overloading -&gt;


"Bernd Jochims" <bjochims@telus .net> wrote in message
news:bjlblf$k5a os$1@ID-51131.news.uni-berlin.de...[color=blue]
>
> "ranjit mario noronha" <noronha@xi.cis .ohio-state.edu> wrote in message
> news:Pine.GSO.4 .40.03090915253 60.271-100000@xi.cis.o hio-state.edu...[color=green]
> >
> > Hi,
> >
> > I am trying to overload the ->operator as follows :
> >
> >
> > class foo {
> > public:
> > int a;
> > int operator->() ;
> >
> > };
> > int foo::operator->() {
> >
> > cout<<"Hullo"<< endl;
> > return 1l;
> > }
> > void main() {
> > foo *f=new foo();
> > cout<<f->a;
> > }
> >
> > However the value of a is printed (0) and the operator function is not
> > called. Does anybody know what mistake I'm making ?
> >
> > Thanks,
> > __
> >
> > Ranjit
> >
> >[/color]
>[/color]
_______________ _______________ _______________ _______________ _______________ _[color=blue]
> ____[color=green]
> >
> > Ranjit Noronha
> > Graduate Research Associate
> > Dept. of Computer and Information Sciences
> > The Ohio State University
> > Phone: (614)477-9900
> > E-mail: noronha@cis.ohi o-state.edu
> >
> >[/color]
>[/color]
_______________ _______________ _______________ _______________ _______________ _[color=blue]
> ____[color=green]
> >
> >///[/color]
> try changing
> foo *f=new foo();
> to
> foo* f= new foo;
>
> also, void main should be int main.
> Bernie
>[/color]
You will see that the data member named "a" is not initialized, returning
garbage. When you forced the initialization to zero, the pointer f returned
0 the value of a..
Using VC7 the following code returns the value of the data member a.

#
#include <iostream>
using namespace std;
class foo {
foo* p;
public:
foo():a(911){}
int a;
foo* operator->() ;
};
foo* foo::operator->() {
cout<<"Hullo"<< endl;
return this;
}
int main() {
foo f;
cout<<(f->a);
}
Hullo
911Press any key to continue

[Unforgiven]

Re: overloading -&gt;

Ron Natalie wrote:[color=blue]
> ranjit mario noronha wrote:[color=green]
>> foo *f=new foo();
>> cout<<f->a;[/color]
>
> This doesn't even attempt to call foo::operator->(). You invoked
> the operator on a foo*. You can't overload on pointers (only
> classes and enums). If you had written:
> foo f;
> cout << f->a;
> things might have been different (of course, with your operator-> it
> still won't compile).[/color]

And in case you're wondering, you can in fact make a call to the operator->
if you have only a pointer to the class: dereference it first.
foo *f = new foo;
(*f)->a;

But you'll still need to fix your other mistakes.

--
Unforgiven

"Most people make generalisations "
Freek de Jonge

[ranjit mario noronha]

Re: overloading -&gt;

[color=blue]
> #include <iostream>
> using namespace std;
> class foo {
> foo* p;
> public:
> foo():a(911){}
> int a;
> foo* operator->() ;
> };
> foo* foo::operator->() {
> cout<<"Hullo"<< endl;
> return this;
> }
> int main() {
> foo f;
> cout<<(f->a);
> }
> Hullo
> 911Press any key to continue
>
>
>[/color]

This is great !! Thanks a lot. But what I really am trying to achieve is
to figure out which element or alternatively the address of the element
which is being accessed. For example for foo->a, the operator should be
able to tell that the element a is being accesses.


Thanks for responding,

Ranjit

[Bernd Jochims]

Re: overloading -&gt;


"ranjit mario noronha" <noronha@kappa. cis.ohio-state.edu> wrote in message
news:Pine.GSO.4 .40.03091016392 70.20023-100000@kappa.ci s.ohio-state.edu...[color=blue]
>[color=green]
> > #include <iostream>
> > using namespace std;
> > class foo {
> > foo* p;
> > public:
> > foo():a(911){}
> > int a;
> > foo* operator->() ;
> > };
> > foo* foo::operator->() {
> > cout<<"Hullo"<< endl;
> > return this;
> > }
> > int main() {
> > foo f;
> > cout<<(f->a);
> > }
> > Hullo
> > 911Press any key to continue
> >
> >
> >[/color]
>
> This is great !! Thanks a lot. But what I really am trying to achieve is
> to figure out which element or alternatively the address of the element
> which is being accessed. For example for foo->a, the operator should be
> able to tell that the element a is being accesses.
>
>
> Thanks for responding,
>
> Ranjit
>[/color]
I'm not sure what you are try to do, but this is what I came up with.
#include <iostream>
#include <vector>
using namespace std;
class foo {
public:
int a,b;
vector<int>vi;
foo(int x = 5763,int n=911):a(n),b(x ){
vi.push_back(50 );
vi.push_back(18 );
vi.push_back(36 );
}
foo* operator->(){
cout<<"Hello from ->"<<endl;
return this;
}
int* operator ->*(int *n)
{
cout<<"hello from ->*"<<endl;
return n;
}
};
int main() {
foo f,g;
cout<<(f->a)<<" address of f.a "<<&f.a<<en dl;
cout<<(f->*(&f.a))<<" address of f.a "<<&f.a<<en dl;
cout<<(f->b)<<" address of f.b "<<&f.b<<en dl;
cout<<(f->*(&f.b))<<" address of f.b "<<&f.b<<en dl;
cout<<(g->*(&g.a))<<" address of g.a "<<&g.a<<en dl;
for (size_t i=0;i<g->vi.size();i+ +)
cout<<(g->*(&g.vi[i]))<<" address of "<<g.vi[i]<<endl;;
}
ouput vc7
Hello from ->
911 address of f.a 0012FEB4
hello from ->*
0012FEB4 address of f.a 0012FEB4
Hello from ->
5763 address of f.b 0012FEB8
hello from ->*
0012FEB8 address of f.b 0012FEB8
hello from ->*
0012FE90 address of g.a 0012FE90
Hello from ->
hello from ->*
002F1130 address of 50
Hello from ->
hello from ->*
002F1134 address of 18
Hello from ->
hello from ->*
002F1138 address of 36
Hello from ->
Press any key to continue

Bernie