How to sort two different objects in an NSArray?

Question

In my NSArray, I have two types of objects, let's say objects of class A and class B. I want to sort these objects, by comparing "startingDate" property of class A to "endingDate" property of class B. And I don't know how to make a comparison based on two different properties of different classes.

What's the best way to do this?

Community · Accepted Answer · 2017-05-23 12:03:33Z

2

Blocks are what you want, take a look at this question: How to sort an NSMutableArray with custom objects in it?

In the blocks example on that question, just have class A use startingDate and class B use ending date.

edited May 23, 2017 at 12:03

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answered Feb 10, 2012 at 15:32

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4 Comments

aslı Over a year ago

Yes I saw that, but all the block methods were comparing the same type of objects. I mean the block's arguments are id obj1 and id obj2, but then they're cast to the same classes. Now that I have two different types of objects, won't casting be a problem? How will I know which argument to cast to my first class?

SushiGrass Jacob Over a year ago

You could check to see what kind of class obj1 and obj2 are by using [[(obj1/obj2) class] isSubclassofClass: [TargetClass class]] and then dealing with appropriately.

aslı Over a year ago

Of course, but it's kind of messy to do so. I'll go that way if there aren't any better options though, thanks.

Jack Lawrence Over a year ago

@Jacob use isKindOfClass since both objects are probably subclasses of NSObject, or they could be subclasses of some other higher level object.

Jack Lawrence · Accepted Answer · 2012-02-16 22:57:04Z

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I would recommend an if/else block and the use of the statement

[myObject isKindOfClass:[ClassA class]]

Which returns a BOOL

If you'd rather (or in conjunction), Apple recommends using

[myObject respondsToSelector:@selector(startingDate)]

Which also returns a BOOL.

Blocks aren't really relevant in this situation; You could use this design pattern inside a plain method implementation, blocks, a for/while loop, or anywhere else.

answered Feb 16, 2012 at 22:57

Jack Lawrence

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