创建一个用计数 N 初始化的 NSArray，所有的对象都是相同的 [英] Creating an NSArray initialized with count N, all of the same object

问题描述

我想创建一个具有相同值的对象(比如 NSNumber 都初始化为 1)的 NSArray，但计数基于另一个变量.除了处理 C 样式数组的初始化器之外，似乎没有办法使用 NSArray 的任何初始化器来做到这一点.

I want to create an NSArray with objects of the same value (say NSNumber all initialized to 1) but the count is based on another variable. There doesn't seem to be a way to do this with any of the intializers for NSArray except for one that deals with C-style array.

不知道有没有捷径可以做到这一点?

Any idea if there is a short way to do this?

这就是我要找的:

NSArray *array = [[NSArray alloc] initWithObject:[NSNumber numberWithInt:0]
                                           count:anIntVariable];

NSNumber 只是这里的一个例子，它本质上可以是任何 NSObject.

NSNumber is just one example here, it could essentially be any NSObject.

推荐答案

我能够为此编写的最紧凑的代码是:

The tightest code I've been able to write for this is:

id numbers[n];
for (int x = 0; x < n; ++x)
    numbers[x] = [NSNumber numberWithInt:0];
id array = [NSArray arrayWithObjects:numbers count:n];

这是可行的，因为您可以使用 Xcode 默认使用的 C99 创建运行时长度确定的 C 数组.

This works because you can create runtime length determined C-arrays with C99 which Xcode uses by default.

如果它们都是相同的值，您也可以使用 memset(尽管转换为 int 是顽皮的):

If they are all the same value, you could also use memset (though the cast to int is naughty):

id numbers[n];
memset(numbers, (int)[NSNumber numberWithInt:0], n);
id array = [NSArray arrayWithObjects:numbers count:n];

如果你知道你需要多少个对象，那么这段代码应该可以工作，尽管我还没有测试过:

If you know how many objects you need, then this code should work, though I haven't tested it:

id array = [NSArray arrayWithObjects:(id[5]){[NSNumber numberWithInt:0]} count:5];

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