Objective-C块和变量

问题描述

我今天开始使用Objective-C块.我写了以下代码:

I've started using Objective-C blocks today. I wrote the following code:

NSArray *array = @[@25, @"abc", @7.2];

void (^print)(NSUInteger index) = ^(NSUInteger index)
{
    NSLog(@"%@", array[index]);
};

for (int n = 0; n < 3; n++)
    print(n);

哪个工作正常.不过，我需要在声明后更改array变量，因此我尝试使用以下代码:

Which works properly. I needed to change the array variable after its declaration, though, so I tried using the following code:

NSArray *array;

void (^print)(NSUInteger index) = ^(NSUInteger index)
{
    NSLog(@"%@", array[index]);
};

array = @[@25, @"abc", @7.2];

for (int n = 0; n < 3; n++)
    print(n);

但是，这不起作用.控制台仅打印(null)三次.为什么它在我的第一段代码中起作用却为什么不起作用?

However, that doesn't work. The console just prints (null) three times. Why is it that this doesn't work, while it did work with my first piece of code?

推荐答案

这是因为该块会按值捕获 和创建该块时的变量(除非您使用__block).

It's because the block captures variables by value and when the block is created (unless you use __block).

您可能想要的是:

NSArray *array = @[@25, @"abc", @7.2];

void (^print)(NSUInteger index) = ^(NSUInteger index)
{
    NSLog(@"%@", array[index]);
};

for (int n = 0; n < 3; n++)
    print(n);

带有__block的示例:

__block NSArray *array;

void (^print)(NSUInteger index) = ^(NSUInteger index)
{
    NSLog(@"%@", array[index]);
};

array = @[@25, @"abc", @7.2];

for (int n = 0; n < 3; n++)
    print(n);

请注意，如果您实际上不需要修改块内的变量并将其反映到外部，则使用__block的效率会降低一些.

Note that it's a little less efficient to use __block if you don't actually need to modify the variable inside the block and have it reflected outside.

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