swift - deinitialize() 与 deallocate()

Question

为什么下面的代码要求我手动使用unsafe_pointer.deinitialize(count: 1)来取消初始化实例？

如果我只使用unsafe_pointer.deallocate()，该实例在我的内存中仍然存在。那么使用deallocate()的目的是什么？

提前抱歉，因为我是第一次尝试学习 MRC。

class Unsafe_Memory {
  var name = "Hello World"
  deinit {
    print("\(type(of: self)) is deinited and deallocated")
  }
}

let unsafe_pointer: UnsafeMutablePointer<Unsafe_Memory> = .allocate(capacity: 1)

unsafe_pointer.pointee = Unsafe_Memory()
dump(unsafe_pointer.pointee)

unsafe_pointer.pointee.name = "Changed"
dump(unsafe_pointer.pointee)

unsafe_pointer.deinitialize(count: 1)
unsafe_pointer.deallocate()

Answer 1

您必须对任何重要类型调用deinitialize。根据Apple Docs ,

A trivial type can be copied bit for bit with no indirection or reference-counting operations. Generally, native Swift types that do not contain strong or weak references or other forms of indirection are trivial, as are imported C structs and enums. Copying memory that contains values of nontrivial types can only be done safely with a typed pointer. Copying bytes directly from nontrivial, in-memory values does not produce valid copies and can only be done by calling a C API, such as memmove().

调用deinitialize首先会取消初始化该内存，但如 docs状态，“调用deinitialize(count:)后，内存未初始化，但仍绑定(bind)到 Pointee 类型”。

之后，您可以通过调用 deallocate() 释放该内存块，类似于 C 中的 free()。deallocate()只能在非平凡类型或未初始化的内存块上调用。

如果类型很简单，则不必调用deinitialize。无论如何，最好调用 deinitialize 以防万一您的类型将来变得不平凡，并且调用它不会花费您任何费用。

您可以尝试运行代码而不调用deallocate。 deinit block 仍将被执行，因为 deinitialize 将运行该 block 。但是，该内存块仍然被占用并且不会被释放，直到您调用deallocate。