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Obj-C 中 For 语句中的多个循环变量

发布于 2024-09-30 07:37:47 字数 179 浏览 1 评论 0原文

有没有办法在 for 语句中使用多个循环变量？以下是我所讨论内容的假设表示：

for (NSString *foo in bar; NSString *bob in example) {}

如果不是，什么是实现此目的的好方法？

谢谢，

科林

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一束光，穿透我孤独的魂 2024-10-07 07:37:47

Joshua 是对的，没有办法用快速枚举语法来做到这一点，但有几种方法（至少）可以实现类似的目标。您喜欢哪种方法取决于您是否试图避免运行两次循环的开销，或者是否试图避免代码中存在两个单独的循环带来的视觉混乱。

假设你有两个 NSString 集合（顺便说一句，它们不需要是 NSArrays）

    NSArray* foo = [NSArray arrayWithObjects:@"bar1", @"bar2", @"bar3", @"bar4", @"bar5", nil];
NSArray* example = [NSArray arrayWithObjects:@"example1", @"example2", nil];

如果你想迭代两个集合的所有元素，你可以执行以下操作：

    NSEnumerator* fooEnumerator = [foo objectEnumerator];
NSEnumerator* exampleEnumerator = [example objectEnumerator];

NSString *bar, *bob;

while ((bar = [fooEnumerator nextObject]) || (bob = [exampleEnumerator nextObject])) {
    if (bar) {
        NSLog(@"%@", bar);
    }
    if (bob) {
        NSLog(@"%@", bob);
    }
}

这将首先迭代 bar 的所有元素，然后遍历示例的所有元素（即它将输出：bar1、bar2、bar3、bar4、bar5、example1、example2）。

如果您想最大限度地减少循环的迭代次数，并且不一定需要遍历两个集合的所有元素，您可以尝试将上面的 while 循环替换为：

while ((bar = [fooEnumerator nextObject]) && (bob = [exampleEnumerator nextObject])) {
    NSLog(@"%@", bar);
    NSLog(@"%@", bob);
}

这将同时循环遍历两个集合，但一旦其中一个集合耗尽了元素（即它将输出：bar1、example1、bar2、example2），

但请注意，循环对性能产生有意义的影响的情况非常罕见，因此如果这都是为了避免运行时成本两个循环可能会通过其他优化得到更好的服务。

Joshua is right that there's no way to do it with fast enumeration syntax but there's a couple of ways (at least) you can achieve something similar. Which approach you prefer depends on whether you're trying avoid the overhead of running the loop twice or whether you're trying to avoid the visual clutter of having two separate loops in your code.

Assuming you have two collections of NSStrings (they don't need to be NSArrays btw)

    NSArray* foo = [NSArray arrayWithObjects:@"bar1", @"bar2", @"bar3", @"bar4", @"bar5", nil];
NSArray* example = [NSArray arrayWithObjects:@"example1", @"example2", nil];

If you're looking to iterate over all the elements of both collections you can do the following:

    NSEnumerator* fooEnumerator = [foo objectEnumerator];
NSEnumerator* exampleEnumerator = [example objectEnumerator];

NSString *bar, *bob;

while ((bar = [fooEnumerator nextObject]) || (bob = [exampleEnumerator nextObject])) {
    if (bar) {
        NSLog(@"%@", bar);
    }
    if (bob) {
        NSLog(@"%@", bob);
    }
}

this will iterate over all of bar's elements first and then over all of example's elements (i.e. it will output: bar1, bar2, bar3, bar4, bar5, example1, example2).

if you want to minimize the number of iterations of the loop and don't necessarily need to run through all the elements of both collections you can try replacing the while loop above with:

while ((bar = [fooEnumerator nextObject]) && (bob = [exampleEnumerator nextObject])) {
    NSLog(@"%@", bar);
    NSLog(@"%@", bob);
}

This will loop through both collections simultaneously but will stop as soon as one of the collections runs out of elements (i.e. it will output: bar1, example1, bar2, example2)

Just as a note though, it's pretty rare that loops have a meaningful impact on performance so if this is all to avoid the runtime cost of two loops you're likely better served by other optimizations.

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ゞ花落谁相伴 2024-10-07 07:37:47

不，无论如何都不使用快速枚举语法。

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奶茶白久 2024-10-07 07:37:47

假设有块，另一种选择：

NSAssert([foo count] == [bar count]);
[foo enumerateUsingBlock: ^(id fooObj, NSUInteger index, BOOL *stopFlag) {
    id barObj = [bar objectAtIndex: index];
    ... do stuff here ...
}];

Assuming blocks, an alternative:

NSAssert([foo count] == [bar count]);
[foo enumerateUsingBlock: ^(id fooObj, NSUInteger index, BOOL *stopFlag) {
    id barObj = [bar objectAtIndex: index];
    ... do stuff here ...
}];

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