| Topic: |
DEVELOP > c-Plus-Plus |
| User: |
"Vasileios" |
| Date: |
27 Nov 2003 06:52:34 AM |
| Object: |
what is the difference ? (pass by reference) |
Hello,
could someone tell me what is the difference between:
1)
int *data;
void doSomething(*data)
{
}
2)
int data;
void doSomething(&data)
{
}
they both pass data by reference right?
Or not?
Thank you
Vasileios
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| User: "friedrich" |
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| Title: Re: what is the difference ? (pass by reference) |
29 Nov 2003 08:50:38 AM |
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1)
int *data;
void doSomething(*data)
This should be:
void doSomething(int *data)
2)
int data;
void doSomething(&data)
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(const 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
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| User: "Jon Bell" |
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| Title: Re: what is the difference ? (pass by reference) |
27 Nov 2003 09:04:25 AM |
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In article <40d3a3ec.0311270452.13ce1851@posting.google.com>,
Vasileios <vasileios@zografos.org> wrote:
1)
int *data;
void doSomething(*data)
{
}
2)
int data;
void doSomething(&data)
{
}
they both pass data by reference right?
Or not?
Not. They're both invalid syntax.
--
Jon Bell <jtbellap8@presby.edu> Presbyterian College
Dept. of Physics and Computer Science Clinton, South Carolina USA
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| User: "Nils Petter Vaskinn" |
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| Title: Re: what is the difference ? (pass by reference) |
27 Nov 2003 07:11:02 AM |
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On Thu, 27 Nov 2003 04:52:34 -0800, Vasileios wrote:
void doSomething(*data)
void doSomething(&data)
they both pass data by reference right?
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
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| User: "Vasileios" |
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| Title: Re: what is the difference ? (pass by reference) |
28 Nov 2003 03:59:16 AM |
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"Nils Petter Vaskinn" <no@spam.for.me.invalid> wrote in message news:<pan.2003.11.27.13.08.39.250657@spam.for.me.invalid>...
On Thu, 27 Nov 2003 04:52:34 -0800, Vasileios wrote:
void doSomething(*data)
void doSomething(&data)
they both pass data by reference right?
No, one passes data by reference the other passes a reference to the data.
Hmm... does this mean that with the first "void doSomething(*data)"
any changes I make inside the function will change the data that is
passed from outside, and the second "void doSomething(&data)" will
only change the data inside the function?
V.Z.
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| User: "Dave OHearn" |
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| Title: Re: what is the difference ? (pass by reference) |
28 Nov 2003 08:08:46 PM |
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(Vasileios) wrote:
Hmm... does this mean that with the first "void doSomething(*data)"
any changes I make inside the function will change the data that is
passed from outside, and the second "void doSomething(&data)" will
only change the data inside the function?
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
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| User: "Kon Tantos" |
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| Title: Re: what is the difference ? (pass by reference) |
29 Nov 2003 01:34:49 PM |
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Vasileios wrote:
"Nils Petter Vaskinn" <no@spam.for.me.invalid> wrote in message news:<pan.2003.11.27.13.08.39.250657@spam.for.me.invalid>...
On Thu, 27 Nov 2003 04:52:34 -0800, Vasileios wrote:
void doSomething(*data)
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 = ...)
void doSomething(&data)
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'.
they both pass data by reference right?
No, one passes data by reference the other passes a reference to the data.
Hmm... does this mean that with the first "void doSomething(*data)"
any changes I make inside the function will change the data that is
passed from outside, and the second "void doSomething(&data)" will
only change the data inside the function?
V.Z.
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| User: "Vasileios Zografos" |
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| Title: Re: what is the difference ? (pass by reference) |
29 Nov 2003 03:25:15 PM |
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Kon Tantos wrote:
Vasileios wrote:
"Nils Petter Vaskinn" <no@spam.for.me.invalid> wrote in message
news:<pan.2003.11.27.13.08.39.250657@spam.for.me.invalid>...
On Thu, 27 Nov 2003 04:52:34 -0800, Vasileios wrote:
void doSomething(*data)
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 = ...)
void doSomething(&data)
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'.
they both pass data by reference right?
No, one passes data by reference the other passes a reference to the
data.
Hmm... does this mean that with the first "void doSomething(*data)"
any changes I make inside the function will change the data that is
passed from outside, and the second "void doSomething(&data)" will
only change the data inside the function?
V.Z.
excellent. thats what I wanted to know.
plain and simple.
Thank you
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| User: "Grumble" |
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| Title: Re: what is the difference ? (pass by reference) |
27 Nov 2003 08:36:46 AM |
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Vasileios wrote:
could someone tell me what is the difference between:
Does this help?
http://www.parashift.com/c++-faq-lite/references.html
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