关于 __bridge、 _bridge_retained 、__bridge_transfer的正确理解

最近看了这样一篇文章http://www.linuxidc.com/Linux/2014-03/97626.htm觉得描述得有点浅，不能简单地说 解不解除ARC的所有权。

所以这里写一下我对这三个方法的理解。

__bridge，不改变对象的持有状态，不retain不release。

<p class="p1"><span class="s1">void</span><span class="s2"> *p = </span><span class="s3">0</span><span class="s2">;</span></p><p class="p1"><span class="s2">{</span></p><p class="p1"><span class="s2">        </span><span class="s1">id</span><span class="s2"> obj = [[</span><span class="s4">NSObject</span><span class="s2"> </span><span class="s5">alloc</span><span class="s2">]</span><span class="s5">init</span><span class="s2">];</span></p><p class="p1"><span class="s2">        </span><span class="s1">void</span><span class="s2"> *p = (</span><span class="s1">__bridge</span><span class="s2"> </span><span class="s1">void</span><span class="s2">*)obj;</span></p><p class="p1"><span class="s2">}</span></p><p class="p2"><span class="s7">//</span><span class="s2">此后访问</span><span class="s7">p</span><span class="s2">出错！（悬垂指针）</span></p>

单纯地转换，这样做容易产生悬垂指针，导致崩溃。

id obj = [[NSObject alloc]init];
void *p = (__bridge_retained void*)obj;

__bridge_retained 使p也持有赋值变量，就是说你在arc环境下也要记得对p进行release。

void *p = (__bridge_retained void *)obj

在no-arc环境下等效于：

id obj = [[NSObject alloc]init];
void *p = obj;
[(id)p retain];

来个例子:obj在{}作用域以外已经释放，p因为进行了retained所以并没有被释放，内存泄露。

void *p = 0;
    {
        id obj = [[NSObject alloc]init];
        p = (__bridge_retained void *)obj;
    }
    NSLog(@"%@",[(__bridge id)p class]);//输出NSObject

__bridge_transfer就是相反

id obj = (__bridge_transfer id)p;

赋值以后马上进行release操作,在no-arc环境下等效于：

id obj = (id)p;
[obj retain];
[(id)p release];