NSMutableArray漏洞内的块（ARC） [英] Blocks inside NSMutableArray leaking (ARC)

问题描述

我有一些块内的操作。此操作只更新 UIImage ，如下所示：

I have some operations that are inside blocks. This operations, only update an UIImage like this:

^(UIImage *image) {
            self.myImage = image;
        }];

我的图片是通过使用 NSURLConnection 。当我从互联网上收到图像时，我称之为 NSMutableArray 内的那个块。到现在为止还挺好。我的问题是，当我有多个具有相同URL的图像，而不是进行多次调用时，我只需在处理连接的类中的 NSMutableArray 中添加一个新块。这样，我进行一次调用，并更新共享URL的多个图像。一切正常，问题是我在泄漏块。我将这样的块添加到 NSMutableArray ：

My image is calculated by accessing the internet with a NSURLConnection. When I receive the image from the internet, I call that block that is inside a NSMutableArray. So far so good. My issue is that when I have multiple images that have the same URL, instead of making multiple calls, I just add a new block to the NSMutableArray inside the class that handles the connection. This way, I make one single call, and update multiple images that share the URL. Everything works ok, the problem is that I am leaking blocks. I add the block like this to the NSMutableArray:

 if( (self = [super init]) ) 
    {
        self.connectionURL=url;
        self.arrayOfBlocks=[NSMutableArray array];
        [arrayOfBlocks addObject:completion];
    }
    return self;

这是初始化连接类的时间。这时我需要在 NSMutableArray 中添加一个新块（我只添加它，就是这样）：

This is when the connection class is initialized. This is when I need to add a new block to the NSMutableArray (I only add it, and that's it):

[arrayOfBlocks addObject:completion];

这是我终于收到回电并开始执行块的时候：

This is when I finally receive the call back, and start executing the blocks:

 for (MyBlock blockToExecute in arrayOfBlocks)
    {
        blockToExecute([UIImage imageWithData:data]);
    }

问题在于这是以某种方式泄漏。我无法抵消使用版本或自动发布，因为我在ARC环境中。那么什么可以解决？

The problem is that this is leaking somehow. I am not able to counteract using releases or auto-releases because I am on ARC environment. So what could be a solution?

推荐答案

这是Apple框架中众所周知的漏洞。这不是因为你使用了积木造成的。

This is a well-known leak in Apple's frameworks. It isn't caused by your use of blocks.

这是关于它的上一个问题和错误报告。

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