What does a square-bracketed index after an NSArray mean? [duplicate]

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What are the details of "Objective-C Literals" mentioned in the Xcode 4.4 release notes? (4 answers)

Closed 8 years ago.

Going through iTunes U Developing iOS 7 Apps for iPhone and iPad and in Lecture 3 slides, on page 120, there's a Quiz question that asks what the following line of code does. Frankly, I'm a bit stumped, and hoped someone could break it down.

cardA.contents = @[cardB.contents,cardC.contents][[cardB match:@[cardC]] ? 1 : 0];

So, I get the first part, cardA.contents = a new array with cardB.contents and cardC.contents in the array. But, next comes (I guess??) an index that returns either 1 or 0 depending if cardB matches an array that includes cardC.

Here's what I don't "get", and maybe it's just a syntax issue.... is what this does?

How is

cardA.contents = @[cardB.contents,cardC.contents][0];

or

cardA.contents = @[cardB.contents,cardC.contents][1];

Valid? Or, did I miss something?

Answer 1

Your understanding is completely correct; you're just missing one bit of syntax. Subscripted access of NSArrays with the array[index] form is part of the "collection literals" feature introduced by Clang a little while back.

It's transformed by the compiler into a call to [array objectAtIndexedSubscript:index].

Answer 2

As usual, in writing this out, it makes sense. It's literally saying, that if cardB matched cardC, use an index of [1] (cardC), if not, use an index of [0] (cardB)

So,

cardA.contents = cardB.contents // if does NOT match
cardA.contents = cardC.contents // if matches

(based on the index).

Duh.. Sorry for a silly question...