无法理解NSError/NSObject指针传递行为 [英] Cannot understand NSError/NSObject pointer passing behavior

问题描述

即使我已阅读 NSError * vs NSError ** 等.

我已经做了一些思考，但仍然有一些疑问.

I've done some thinking and still got some questions.

这是我写的:

NSError *error = [NSError errorWithDomain:@"before" code:0 userInfo:nil];
NSLog(@"outside error address: %p", &error];
[self doSomethingWithObj:nil error:&error];

为了测试上述NSError方法，我这样写:

In order to test the above NSError method, I wrote this:

- (id)doSomethingWithObj:(NSObject *)obj error:(NSError *__autoreleasing *)error
{
    NSLog(@"inside error address: %p", error);
    id object = obj;
    if (object != nil)
    {
        return object;
    }
    else
    {
        NSError *tmp = [NSError errorWithDomain:@"after" code:0 userInfo:nil];
        *error = tmp;
        return nil;
    }
}

但是我发现两个日志记录地址不同.为什么会这样?

But I found that the two logging addresses are different. Why is that?

2016-08-19 19:00:16.582 Test[4548:339654] outside error address: 0x7fff5b3e6a58
2016-08-19 19:00:16.583 Test[4548:339654] inside error address: 0x7fff5b3e6a50

它们不应该是相同的，因为那只是一个简单的值副本?如果它们应该不同，那么指针指针如何最终指向同一NSError实例?

Shouldn't they be the same since that was just a simple value copy? If they should be different, how can pointer to pointer end up pointing to the same NSError instance?

推荐答案

调用方中的变量的类型为NSError*.该地址的类型为NSError* *.函数期望NSError* __autoreleasing *.因此，编译器将创建类型为NSError* __autoreleasing的隐藏变量，将NSError*复制到调用之前的隐藏变量中，并在调用之后将其复制回以获得__autoreleasing的语义.

The variable in the caller has type NSError*. The address has type NSError* *. The function expect NSError* __autoreleasing *. Therefore the compiler creates a hidden variable of type NSError* __autoreleasing, copies the NSError* into the hidden variable before the call, and copies it back after the call to get the semantics of __autoreleasing right.

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