返回结构在 Objective-C 中究竟是如何工作的? [英] How exactly does returning structs work in Objective-C?

问题描述

假设我有以下函数返回一个 CLLocationCoordinate2D，它只是一个 struct，它有两个 double(名为 经度和纬度):

Say I have the following function that returns a CLLocationCoordinate2D, which is just a struct that has two doubles (named longitude and latitude) in it:

- (CLLocationCoordinate2D)getCoordinateForHouse:(House *)house
{
    CLLocationCoordinate2D coordToReturn;
    coordToReturn.latitude = // get house's latitude somehow
    coordToReturn.longitude = // get house's longitude somehow

    return coordToReturn;
}

我可以基本上像对待任何其他原始类型一样对待这个 struct 吗?例如，如果我在其他地方的代码中调用上面的函数，如下所示:

Can I basically treat this struct just like any other primitive type? For instance, if I call the function above in code somewhere else like this:

CLLocationCoordinate2D houseCoord = 
       [someClassThatTheAboveFunctionIsDefinedIn getCoordinatesForHouse:myHouse];

从函数返回的值被简单地复制到 houseCoord(就像任何其他原语一样)，对吗?我不必担心 CLLocationCoordinate2D 在其他地方被破坏?

The value that was returned from the function is simply copied into houseCoord (just like any other primitive would act), right? I don't have to worry about the CLLocationCoordinate2D ever being destroyed elsewhere?

现在这对我来说似乎很明显，但我只需要确认.

Seems obvious to me now that this is probably the case, but I just need confirmation.

推荐答案

这是 Objective-C 直接从 C 继承其行为的领域——分配结构会导致浅拷贝.所以如果你在结构中有任何指针，那么你将复制指针，而不是指向的东西.在 C 代码中，您需要将其纳入有关所有权的临时规则.在 Objective-C 中，自动引用计数编译器无法处理对结构(或者，我认为，联合)内对象的引用，因此在任何情况下都不要太执着于这个想法是明智的.

This is an area where Objective-C inherits its behaviour directly from C — assigning structs causes a shallow copy. So if you have any pointers inside the structs then you'll copy the pointer, not the thing pointed to. In C code you need to factor that in to your adhoc rules about ownership. In Objective-C the automatic reference counting compiler isn't capable of dealing with references to objects within structs (or, I think, unions) so it's smart not to become too attached to the idea in any event.

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