Objective-C传递块作为参数 [英] Objective-C pass block as parameter

问题描述

如何将Block传递给Function/Method?

我尝试- (void)someFunc:(__Block)someBlock无济于事.

即. Block的类型是什么?

推荐答案

块的类型取决于其自变量和返回类型.在一般情况下，块类型的声明方式与函数指针类型的声明方式相同，只是将*替换为^.将块传递给方法的一种方法如下:

The type of a block varies depending on its arguments and its return type. In the general case, block types are declared the same way function pointer types are, but replacing the * with a ^. One way to pass a block to a method is as follows:

- (void)iterateWidgets:(void (^)(id, int))iteratorBlock;

但是您可以看到，这很混乱.您可以改为使用typedef来使块类型更整洁:

But as you can see, that's messy. You can instead use a typedef to make block types cleaner:

typedef void (^ IteratorBlock)(id, int);

然后将该块传递给类似这样的方法:

And then pass that block to a method like so:

- (void)iterateWidgets:(IteratorBlock)iteratorBlock;

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