Block_copy是递归的吗？ [英] Is Block_copy recursive?

问题描述

我有一些基本上归结为这样的代码：

I have some code that essentially boils down to this:

-(void)doSomethingWithBlock:(BlockTypedef)block
{
    [Foo doSomethingElseWithBlock:^() {
        block();
    }];
}

Foo doSomethingElseWithBlock：在它接收的块上调用 Block_copy 和 Block_release 。这在外部范围也是必要的，或者内部 Block_copy 处理这个？

Foo doSomethingElseWithBlock: calls Block_copy and Block_release on the block that it receives. Is this also necessary at the outer scope, or will the inner Block_copy handle this?

推荐答案

我引用阻止编程主题指南在Apple的开发者文档网站：

I quote the Blocks Programming Topics guide on Apple's developer documentation site:

复制块时，任何如果需要，则复制对该块内的其它块的引用 - 可以复制（从顶部）整个树。

When you copy a block, any references to other blocks from within that block are copied if necessary—an entire tree may be copied (from the top). If you have block variables and you reference a block from within the block, that block will be copied.

当你复制一个基于堆栈的块时，你会得到一个新的块。但是，如果复制基于堆的块，则只需增加该块的保留计数，并将其返回为复制函数或方法的返回值。

When you copy a stack-based block, you get a new block. If you copy a heap-based block, however, you simply increment the retain count of that block and get it back as the returned value of the copy function or method.

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