目标 C 相当于 PHP 范围函数? [英] Objective C equivalent of PHP range function?

问题描述

我想快速创建一个从 0 到 10000 的数字数组，但步进增量为 500，对于如下所示的数组:

I want to quickly create an array of numbers from 0 to 10000 but with stepping increments of 500, for an array that looks like this:

@[@0, @500, @1000, ... @10000]

在 PHP 中，它内置于 range 函数，你可以像这样生成它:range(0, 10000, 500)

in PHP, this is built in to the range function, you could generate it like this: range(0, 10000, 500)

我并不是真的在寻找答案，比如循环遍历一些初始数组或数字超过 0-10000 并使用模数或其他东西，我只是想知道是否有内置函数或常用速记来执行此操作

I'm not really looking for answers like looping over some initial array or numerically over 0-10000 and using modulus or something, I just want to know if there's a built in function or common shorthand for doing this

推荐答案

Objective-C 中没有类似的内置函数.正如您所暗示的，最好的方法是循环如下:

There is no built-in function like that in Objective-C. As you hinted, the best way to do it is by looping as follows:

NSMutableArray *array = [[NSMutableArray alloc] init];
for (NSInteger i = 0; i <= 10000; i += 500) {
    [array addObject:@(i)];
}

如果您想创建一个与 PHP 范围函数类似的方法:返回从 start 到 end 的元素数组，包括.

If you want to make a method that works similarly to the PHP range function: Returns an array of elements from start to end, inclusive.

这样做:

+ (NSArray *)rangeWithStart:(NSInteger)start end:(NSInteger)end step:(NSInteger)step {
    NSMutableArray *array = [[NSMutableArray alloc] init];
    for (NSInteger i = start; i <= end; i += step) {
        [array addObject:@(i)];
    }
    return [array copy];
}

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