Cannot understand NSError/NSObject pointer passing behavior

Question

I am now confused by pointer to pointer even though I've read Why does NSError need double indirection? (pointer to a pointer) and NSError * vs NSError ** and much more.

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

Here I wrote this:

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

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;
    }
}

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

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?

Answer 1

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.