NSArray之后的方括号索引是什么意思？ [英] What does a square-bracketed index after an NSArray mean?

问题描述

通过iTunes U开发适用于iPhone和iPad的iOS 7应用程序，以及第120页的Lecture 3幻灯片，有一个测验问题，询问下面的代码行是什么。坦白地说，我有点累了，希望有人能打破它。

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];

所以，我得到第一部分， cardA.contents = 在数组中使用 cardB.contents 和 cardC.contents 的新数组。但是，接下来（我想是）一个索引返回1或0，取决于 cardB 是否匹配包含 cardC 。

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.

这里是我不get，也许只是一个语法问题....这是什么？

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

如何

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

或

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

有效？

推荐答案

你的理解是完全正确的;你只是缺少一点语法。通过 NSArray 的下标访问> 数组[index] 表单是Clang稍微介绍的集合文字功能的一部分。

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.

它由编译器转换为 [array objectAtIndexedSubscript：index] 。

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