将块应用于 NSArray 的所有对象

发布于 2024-10-20 10:27:29 字数 963 浏览 0 评论 0原文

有没有办法将块应用于数组的所有对象？

假设我有一个令人惊叹的块：

void (^myAmazingBlock)(NSNumber *) = ^(NSNumber *aFooNumber) {
    NSLog(@"Log message from an Amazing Block: %@", aFooNumber);
};

要将我的块应用到数组的所有对象，这是可行的：

for (NSNumber *aNumber in myArray) {
    myAmazingBlock(aNumber);
}

是否可以摆脱快速枚举循环，并在精神上有类似的东西：

[myArray valueForKeyPath:@"@distinctUnionOfObjects.^myAmazingBlock"]; // This code doesn't work. It's just to show the style of what I'm trying to write.

我没有任何具体用途要牢记；我只是好奇其他可能的写法。

编辑：

可以使用enumerateObjectsUsingBlock：（谢谢paulbailey）。
然后，您必须像这样声明您的块：

void (^myAmazingBlock2)(id, NSUInteger, BOOL *) = ^(id aFooNumber, NSUInteger idx, BOOL *stop) {
    NSLog(@"Amazing BLOCK 2 %@", aFooNumber);
};

[myArray enumerateObjectsUsingBlock:myAmazingBlock2];

不幸的是，快速枚举循环仍然是最具可读性的解决方案。

原文

Is there a way to apply a block to all the objects of an array?

Let's say I have an amazing block:

void (^myAmazingBlock)(NSNumber *) = ^(NSNumber *aFooNumber) {
    NSLog(@"Log message from an Amazing Block: %@", aFooNumber);
};

To apply my block to all the objects of my array, this works:

for (NSNumber *aNumber in myArray) {
    myAmazingBlock(aNumber);
}

Is it possible to get rid of the fast enumeration loop, and have something similar in the spirit to:

[myArray valueForKeyPath:@"@distinctUnionOfObjects.^myAmazingBlock"]; // This code doesn't work. It's just to show the style of what I'm trying to write.

I don't have any specific use in mind ; I was just being curious of the other possible ways to write this.

Edit:

It's possible to use enumerateObjectsUsingBlock: (Thank you paulbailey).
You then have to declare your block like this:

void (^myAmazingBlock2)(id, NSUInteger, BOOL *) = ^(id aFooNumber, NSUInteger idx, BOOL *stop) {
    NSLog(@"Amazing BLOCK 2 %@", aFooNumber);
};

[myArray enumerateObjectsUsingBlock:myAmazingBlock2];

Unfortunately, the fast enumeration loop is still the most readable solution.

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