swift - 从 `self` 走私 `deinit`

Question

如果我走私会怎样self出我的 deinit ，通过将其分配给某个外部强引用？下面的这段代码显然格式不正确:

class C: CustomStringConvertible {
    let s = "abc"

    var description: String {
        return "C(id: \(ObjectIdentifier(self)), s: \(s))"
    }

    deinit {
        print("deinit")
        globalObject = self
    }
}

var globalObject: C!

do {
    let localObject = C()
    print("localObject: \(localObject)")
    print("end of `do`")
}

print("globalObject: \(globalObject!)")

您不能仅仅从 deinit 的中间“改变主意”来取消对象的初始化。 .但有趣的是，这段代码是不确定的，它有时会成功完成，打印:

localObject: C(id: ObjectIdentifier(0x00007f9063f00960), s: abc)
end of `do`
deinit
globalObject: C(id: ObjectIdentifier(0x00007f9063f00960), s: abc)

我正在使用 Code Runner 运行它，它只是使用 swiftc 运行单个文件 Swift 脚本。 .所以这里没有意外的 Playground 拥有的引用。

非确定性从何而来？

Answer 1

这不是答案，但评论太长了:
有趣的是，我只是附加了:

for _ in 1...1000000 {
    print("intermediate: \(globalObject!)")
}
print("globalObject: \(globalObject!)")

进而:

swiftc test.swift
for ((i = 0; i < 10000; i++)); do
    ./test  | fgrep globalObject
done

我得到的是(只有 10000 次运行中的一次):

Fatal error: Object was retained too many timesa(44854,0x10a92f5c0) malloc: Incorrect checksum for freed object 0x7fcc1ec02b98: probably modified after being freed.

Corrupt value: 0x7ffffffe00000000

test(44854,0x10a92f5c0) malloc: *** set a breakpoint in malloc_error_break to debug

globalObject: C(id: ObjectIdentifier(0x00007f7f452006c0), s: abc)

test(61212,0x114d7e5c0) malloc: Incorrect checksum for freed object 0x7fe063e00008: probably modified after being freed.

Corrupt value: 0x4ffffffe00000000

test(61212,0x114d7e5c0) malloc: *** set a breakpoint in malloc_error_break to debug

下一次运行:

test(7016,0x10682d5c0) malloc: tiny_free_list_remove_ptr: Internal invariant broken (next ptr of prev): ptr=0x7fb2e0805fd0, prev_next=0x7f92e0805fd0

这似乎是一些竞争条件，但我不知道谁在这里产生线程。
顺便提一句:

swiftc -version
Apple Swift version 5.0.1 (swiftlang-1001.0.82.4 clang-1001.0.46.5)
Target: x86_64-apple-darwin18.7.0