what is the difference ? (pass by reference)

[Vasileios]

Hello,

could someone tell me what is the difference between:


1)

int *data;

void doSomething(*da ta)
{

}



2)
int data;
void doSomething(&da ta)
{

}



they both pass data by reference right?
Or not?

Thank you
Vasileios

[Nils Petter Vaskinn]

Re: what is the difference ? (pass by reference)

On Thu, 27 Nov 2003 04:52:34 -0800, Vasileios wrote:

[color=blue]
> void doSomething(*da ta)
> void doSomething(&da ta)[/color]
[color=blue]
> they both pass data by reference right?[/color]

No, one passes data by reference the other passes a reference to the data.

--
NPV

"the large print giveth, and the small print taketh away"
Tom Waits - Step right up

[Grumble]

Re: what is the difference ? (pass by reference)

Vasileios wrote:[color=blue]
> could someone tell me what is the difference between:[/color]

Does this help?

http://www.parashift.com/c++-faq-lite/references.html

[Jon Bell]

Re: what is the difference ? (pass by reference)

In article <40d3a3ec.03112 70452.13ce1851@ posting.google. com>,
Vasileios <vasileios@zogr afos.org> wrote:[color=blue]
>
>1)
>
>int *data;
>
>void doSomething(*da ta)
>{
>
>}
>
>
>
>2)
>int data;
>void doSomething(&da ta)
>{
>
>}
>
>
>
>they both pass data by reference right?
>Or not?[/color]

Not. They're both invalid syntax.

--
Jon Bell <jtbellap8@pres by.edu> Presbyterian College
Dept. of Physics and Computer Science Clinton, South Carolina USA

[Vasileios]

Re: what is the difference ? (pass by reference)

"Nils Petter Vaskinn" <no@spam.for.me .invalid> wrote in message news:<pan.2003. 11.27.13.08.39. 250657@spam.for .me.invalid>...[color=blue]
> On Thu, 27 Nov 2003 04:52:34 -0800, Vasileios wrote:
>
>[color=green]
> > void doSomething(*da ta)
> > void doSomething(&da ta)[/color]
>[color=green]
> > they both pass data by reference right?[/color]
>
> No, one passes data by reference the other passes a reference to the data.[/color]


Hmm... does this mean that with the first "void doSomething(*da ta)"
any changes I make inside the function will change the data that is
passed from outside, and the second "void doSomething(&da ta)" will
only change the data inside the function?


V.Z.

[Dave O'Hearn]

Re: what is the difference ? (pass by reference)

vasileios@zogra fos.org (Vasileios) wrote:[color=blue]
> Hmm... does this mean that with the first "void doSomething(*da ta)"
> any changes I make inside the function will change the data that is
> passed from outside, and the second "void doSomething(&da ta)" will
> only change the data inside the function?[/color]

It would help if you gave some info on how long you have been using
C++. Are you familiar with pointers and trying to understand how
references are different, or are you new to all of this stuff at once?

--
Dave O'Hearn

[friedrich]

Re: what is the difference ? (pass by reference)

> 1)[color=blue]
>
> int *data;
>
> void doSomething(*da ta)[/color]

This should be:

void doSomething(int *data)

[color=blue]
> 2)
> int data;
> void doSomething(&da ta)[/color]

This should be:

void doSomething(int &data)

In the first example you would have to use a pointer-to-int as the
function argument, in the second, you could use an int as an argument
and the function would automatically use a reference. If you used
const:

void doSomething(con st int &data)

that would protect the int from being changed outside of the function.

Your function name looks like a Java style name. Are you a Java
programmer?

-FA

[Kon Tantos]

Re: what is the difference ? (pass by reference)



Vasileios wrote:[color=blue]
> "Nils Petter Vaskinn" <no@spam.for.me .invalid> wrote in message news:<pan.2003. 11.27.13.08.39. 250657@spam.for .me.invalid>...
>[color=green]
>>On Thu, 27 Nov 2003 04:52:34 -0800, Vasileios wrote:
>>
>>
>>[color=darkred]
>>>void doSomething(*da ta)[/color][/color][/color]
this passes a copy of a reference (called a pointer). Your function can change 'data'
without affecting the copy in the calling code. You can dereference 'data' to change
data in the calling code (*data = ...)
[color=blue][color=green][color=darkred]
>>>void doSomething(&da ta)[/color][/color][/color]
this passes a reference to data in the calling code. When you access data you are
_actually_ working with the outside data. So data = ... directly changes whatever
data is 'refering to'.
[color=blue][color=green]
>>
>>
>>[color=darkred]
>>>they both pass data by reference right?[/color]
>>
>>No, one passes data by reference the other passes a reference to the data.[/color]
>
>
>
> Hmm... does this mean that with the first "void doSomething(*da ta)"
> any changes I make inside the function will change the data that is
> passed from outside, and the second "void doSomething(&da ta)" will
> only change the data inside the function?
>
>
> V.Z.[/color]

[Vasileios Zografos]

Re: what is the difference ? (pass by reference)

Kon Tantos wrote:[color=blue]
>
>
> Vasileios wrote:
>[color=green]
>> "Nils Petter Vaskinn" <no@spam.for.me .invalid> wrote in message
>> news:<pan.2003. 11.27.13.08.39. 250657@spam.for .me.invalid>...
>>[color=darkred]
>>> On Thu, 27 Nov 2003 04:52:34 -0800, Vasileios wrote:
>>>
>>>
>>>
>>>> void doSomething(*da ta)[/color][/color]
>
> this passes a copy of a reference (called a pointer). Your function can
> change 'data' without affecting the copy in the calling code. You can
> dereference 'data' to change data in the calling code (*data = ...)
>[color=green][color=darkred]
>>>> void doSomething(&da ta)[/color][/color]
>
> this passes a reference to data in the calling code. When you access
> data you are _actually_ working with the outside data. So data = ...
> directly changes whatever data is 'refering to'.
>[color=green][color=darkred]
>>>
>>>
>>>
>>>> they both pass data by reference right?
>>>
>>>
>>> No, one passes data by reference the other passes a reference to the
>>> data.[/color]
>>
>>
>>
>>
>> Hmm... does this mean that with the first "void doSomething(*da ta)"
>> any changes I make inside the function will change the data that is
>> passed from outside, and the second "void doSomething(&da ta)" will
>> only change the data inside the function?
>>
>>
>> V.Z.[/color]
>
>[/color]

excellent. thats what I wanted to know.
plain and simple.

Thank you