What does this mean?

[Protoman]

What does this do?

int& fn(int& arg){arg++;retu rn arg;}
//later
int y;
fn(y)=5;

Why can you do it? What do you use it for? Thanks!!!

[Dan Cernat]

Re: What does this mean?

See below

"Protoman" <Protoman2050@g mail.com> wrote in message
news:1134267381 .114631.110770@ g43g2000cwa.goo glegroups.com.. .[color=blue]
> What does this do?
>
> int& fn(int& arg){arg++;retu rn arg;}
> //later
> int y;
> fn(y)=5; <-- use of uninitialised variable y[/color]
according to the body of fn, you may get an overflow. otherwise the
statement above will initialise y with 5
[color=blue]
>
> Why can you do it?[/color]
Why not?
What do you use it for?
I would say if you want to change the value of a static variable inside fn
between calls to fn
example (untested)

int& fn(int arg)
{
static int persist = 2;

// so some work here

return persist;
}

int main()
{
fn(1); // persist == 2 after this call
fn(2) = 7; // persist will be 7 after this
fn(3); // persist is still 7

return 0;
}

Thanks!!![color=blue]
>[/color]

Dan

[Rolf Magnus]

Re: What does this mean?

Dan Cernat wrote:
[color=blue]
> "Protoman" <Protoman2050@g mail.com> wrote in message
> news:1134267381 .114631.110770@ g43g2000cwa.goo glegroups.com.. .[color=green]
>> What does this do?
>>
>> int& fn(int& arg){arg++;retu rn arg;}
>> //later
>> int y;
>> fn(y)=5; <-- use of uninitialised variable y[/color]
> according to the body of fn, you may get an overflow. otherwise the
> statement above will initialise y with 5[/color]

Actually, the behavior will be undefined, because the value of an
uninitialized variable is used (within the function). That happens before
the assignment.