The out and ref Parameter in C#

Question

3.50/5 (2 votes)

See more:

what is the difference between them?
when or where we should use OUT and REF?
Programming example?
RealWorld Example
..
thank you.
Positive discussion is appreciated.

Posted 13-Dec-11 4:55am

awaisdar

Add a Solution

Comments

Philippe Mori 13-Dec-11 20:19pm

By the way, the compiler documentation is quite adequate also:
out parameter modifier and ref (C# Reference)

5 solutions

Solution 5

using ref keyword requires that the variable be initialized before being passed
out-no need to initialize before passing to function

Posted 13-Jan-13 8:27am

mishraamit09

Comments

Manfred Rudolf Bihy 13-Jan-13 14:49pm

What has your solution added that all the others didn't already say?
Reported as abuse!

Jibesh 13-Jan-13 16:27pm

also why you answered for the question which a year old and its already solved.

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Sergey Alexandrovich Kryukov · Accepted Answer · 2011-12-13T05:03:00

Solution 1

They are almost the same and their low-level implementations are identical; the only difference is the syntax requirements to the implementation of the method.

If the argument is "out", the compiler enforces assignment of a new value to the parameter. More exactly, if a compiler can see the situation where this parameter is left unassigned, it generates a compilation error. This is a powerful fool-proof feature.

The "ref" argument impose more relaxed rules which is adequate to "in-out" behavior. Modification of the value of this parameter is not required. If the value is not modifies on return, it's fine, the value will remain the same as before the call.

[EDIT] I forgot to mention that "out" variable does not have to be initialized. The requirement I mentioned above allows to consider such variable as initialized after the call and guarantee initialization. Of course "ref" parameter is required to be initialized. Credit to Monjurul.

—SA

Posted 13-Dec-11 5:03am

Sergey Alexandrovich Kryukov

Updated 13-Dec-11 7:07am

v5

Comments

fjdiewornncalwe 13-Dec-11 12:17pm

Well explained. +5

Sergey Alexandrovich Kryukov 13-Dec-11 12:28pm

It was easy, in contrast to many other cases. :-)
Thank you, Marcus.
--SA

Abhinav S 13-Dec-11 12:34pm

My 5!

Sergey Alexandrovich Kryukov 13-Dec-11 12:35pm

Thank you, Abhinav.
--SA

Monjurul Habib 13-Dec-11 12:56pm

5!

Sergey Alexandrovich Kryukov 13-Dec-11 13:04pm

Thank you, Monjurul; and thanks again for helping to correct this answer.
--SA

Monjurul Habib 13-Dec-11 13:06pm

anytime, my pleasure

RaviRanjanKr 13-Dec-11 16:10pm

My 5 for nice answer :)

Sergey Alexandrovich Kryukov 13-Dec-11 16:17pm

Thank you, Ravi.
--SA

Wonde Tadesse 13-Dec-11 19:38pm

5+

Sergey Alexandrovich Kryukov 13-Dec-11 20:16pm

Thank you, Wonde.
--SA

[no name] 13-Dec-11 20:50pm

5!

Sergey Alexandrovich Kryukov 13-Dec-11 21:26pm

Thank you, Eduard.
--SA

RakBallal 2-Mar-13 3:58am

Well explained..thank U

Sergey Alexandrovich Kryukov 2-Mar-13 20:09pm

Thank you.
—SA

Monjurul Habib · Accepted Answer · 2011-12-13T05:47:00

Solution 2

Both are pretty similar - the only difference is that a variable you pass as an out parameter doesn't need to be initialized, and the method using the out parameter has to set it to something.

C#

int a;
Item(out a); // OK

int b;
Item(ref b); // Not Ok

Ref parameters are for data that might be modified, out parameters are for data that's an additional output for the function (eg int.TryParse) that are already using the return value for something.

For more detail information please visit the following link:
The out and ref Paramerter in C#
What is the difference between out and ref in C#?

Posted 13-Dec-11 5:47am

Monjurul Habib

Updated 13-Dec-11 5:52am

v3

Comments

Sergey Alexandrovich Kryukov 13-Dec-11 12:32pm

Oh gosh, I forget to mention about initialization, but now I fixed my answer, with some extra explanations.
Thank you and my 5. My credit for your answer is in my updated solution.
--SA

Monjurul Habib 13-Dec-11 12:55pm

thank you

Abhinav S 13-Dec-11 12:34pm

Useful to the OP. 5.

Monjurul Habib 13-Dec-11 12:55pm

thank you

Uday P.Singh 13-Dec-11 12:49pm

5ed :)

Monjurul Habib 13-Dec-11 12:55pm

thank you

RaviRanjanKr 13-Dec-11 16:10pm

My 5!

Monjurul Habib 13-Dec-11 22:44pm

thank you

Wonde Tadesse 13-Dec-11 18:09pm

5+

Monjurul Habib 13-Dec-11 22:44pm

thank you

[no name] 13-Dec-11 20:50pm

5!

Monjurul Habib 13-Dec-11 22:45pm

thank you

Abhinav S · Accepted Answer · 2011-12-13T06:34:00

Solution 3

I searched and came across this question[^] asked a few months ago. Should help you out.

Posted 13-Dec-11 6:34am

Abhinav S

Comments

Monjurul Habib 13-Dec-11 12:57pm

yes :) 5!

Abhinav S 13-Dec-11 13:03pm

Thanks. :)

[no name] 13-Dec-11 20:51pm

5!

Abhinav S 14-Dec-11 0:15am

Thank you.

idle63 · Accepted Answer · 2011-12-13T09:55:00

Solution 4

Only to become a feeling: Before a long time with old programming languages there was something like passing parameters by reference (often misused to avoid overhead when passing by value). But nobody checked really whether the parameter(s) has been set by the called method.

So with the “out parameter” some unknowns are removed at compile time, because it guarantees that the called method has to assign something to these parameters.

And by the way: Removing uncertainty means Information ;)

Regards

Posted 13-Dec-11 9:55am

idle63

Updated 13-Dec-11 11:11am

v6

Comments

Sergey Alexandrovich Kryukov 13-Dec-11 20:17pm

Ha! Well said. My 5.
--SA

[no name] 23-Dec-11 12:41pm

thx. Regards

[no name] 13-Dec-11 20:51pm

5!

[no name] 23-Dec-11 12:41pm

thx. Regards