passing out parameter

Question

I wrote a method with an out parameter:

-(NSString *)messageDecryption:(NSString *)receivedMessage outParam:(out)messageCondent
{   
    messageCondent = [receivedMessage substringFromIndex:2];
    return [receivedMessage substringToIndex:1];
}

Then I passed the param like this:

NSString *messageCondent;
NSString *mode = [myclassobject messageDecryption:message outParam:messageCondent];

However, there is a problem. The out parameter value is not being set properly. Can any one help me to do this correctly?

I've never actually seen this `out` method used before. I think most people simply implement value by reference. (Check the correct answer on this page: https://discussions.apple.com/thread/1502936?threadID=1502936) — peterp, May 16 '11 at 17:28
Take into consideration that [pass by reference in Cocoa/iOS is largely limited to `NSError**`](http://stackoverflow.com/questions/3331791/arguments-by-reference-in-objective-c/3332062#3332062). If you need to return more than one value at a time, that begs for a structure or, more often, a class. — albertamg, May 16 '11 at 17:36
`out` is a keyword in Objective-C that was related to Distributed Objects. It isn't used much anymore. — bbum, May 16 '11 at 19:56

score 28 · Accepted Answer · answered May 16 '11 at 17:29

28

Create the method to accept a pointer to the object.

-(NSString *)messageDecryption:(NSString *)receivedMessage outParam:(NSString**)messageCondent
{   
    *messageCondent = [receivedMessage substringFromIndex:2];
    return [receivedMessage substringToIndex:1];

}

Pass in the reference to the local object.

NSString *messageCondent = nil;
NSString *mode = [myclassobject messageDecryption:message outParam:&messageCondent];

answered May 16 '11 at 17:29

tidwall

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and for giving an explicit type to the parameter. – jscs May 16 '11 at 18:30
Note that use of out parameters in Objective-C is exceedingly rare; pretty much limited only `NSError**`... – bbum May 16 '11 at 19:56
1

is there any major problm with out param? – Vipin May 16 '11 at 20:36

Jacob Relkin · Answer 2 · 2011-05-16T18:43:44.167

11

An "out parameter" is by definition a pointer to a pointer.

Your method should look like this:

-(NSString *)messageDecryption:(NSString *)receivedMessage outParam:(NSString **)messageCondent
{   
    *messageCondent = [receivedMessage substringFromIndex:2];
    return [receivedMessage substringToIndex:1];
}

This dereferences the passed-in pointer to get at the actual object reference and then assigns that to whatever [receivedMessage substringFromIndex:2] returns.

Invoking this method is quite simple:

NSString *messageCondent = nil;
NSString *mode = [myclassobject messageDecryption:message outParam:&messageCondent];

edited May 16 '11 at 18:43

answered May 16 '11 at 17:29

Jacob Relkin

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No, an "out parameter" is a pointer to a *value*. That the value is a pointer is only true if the value is an object. But it could just as well be an `int` – Thomas Tempelmann Sep 04 '20 at 10:04

passing out parameter

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