将值输出到 swift 中的 typedef 指针 [英] outputting a value to a typedef pointer in swift

问题描述

我几乎可以肯定这个标题不正确，但这里是...

I'm almost certain the title of this isn't correct but here goes...

我正在桥接到一个 Objective-C 类来设置一个 typedef.桥已设置好，我可以正确声明 typedef var.

I'm bridging to an Objective-C class to set a typedef. The bridge is set up and I'm able to declare the typedef var correctly.

在 Objective-C 中，我还从同一个类中调用了一个方法，该方法在调用时会向变量 TestHandle 输出一个值.

In Objective-C I also called a method from the same class that, when called, output a value to the variable TestHandle.

var TestHandle : TESTHANDLE
TestInit(&TestHandle)

当我使用 Swift 5 尝试此操作时，出现此错误:

When I try this using Swift 5 I get this error:

Cannot convert value of type 'inout TESTHANDLE' (aka 'inout UnsafeMutableRawPointer') to expected argument type 'UnsafeMutablePointer<TESTHANDLE?>?' (aka 'Optional<UnsafeMutablePointer<Optional<UnsafeMutableRawPointer>>>')

有什么指点吗?

推荐答案

一些观察:

• TESTHANDLE 似乎是 UnsafeMutableRawPointer
的别名
• &testHandle 正在获取 testHandle 的引用(指向位置的指针)，生成 inout UnsafeMutableRawPointer 类型的值/li>
• 正如错误所说，您的 TestInit 函数采用 UnsafeMutablePointer<TESTHANDLE?>? 类型的变量，又名 Optional

• TESTHANDLE appears to be an alias for UnsafeMutableRawPointer
• &testHandle is taking a reference (a pointer to the location) of the testHandle, producing a value of type inout UnsafeMutableRawPointer
• As the error says, your TestInit function takes a variable of type UnsafeMutablePointer<TESTHANDLE?>?, a.k.a. Optional<UnsafeMutablePointer<Optional<UnsafeMutableRawPointer>>>

Swift 有一些关于 & 如何自动桥接各种指针类型的规则，但坦率地说，我不太了解它们.

Swift has some rules about how & automatically bridges to the various pointer types, but to be frank, I don't understand them very well.

据我所知，Swift 指针类型不能表示 nil (0x000...000).要做到这一点，它们需要包含在一个可选项中.所以当你看到类型

As far as I know, the Swift pointer types cannot represent nil (0x000...000). To do that, they need to be wrapped within an optional. So when you see the type

Optional<UnsafeMutablePointer<Optional<UnsafeMutableRawPointer>>>

它实际上是两个语义"部分:

It's actually two "semantic" parts:

Optional<UnsafeMutablePointer<    Optional<UnsafeMutableRawPointer>    >>
↳ A nullable pointer to ...       ↳ ... something that's a nullable pointer of unspecified (void) type

您收到错误的原因是 &testHandle 只能将您的 UnsafeMutableRawPointer 桥接到 Optional<UnsafeMutablePointer<UnsafeMutableRawPointer>>code>，但不是必需的 Optional>>(区别在于缺少内部"可空性层).要解决此问题，请自行设置 testHandle 可选:

The reason you're getting your error is because &testHandle can only bridge your UnsafeMutableRawPointer to a Optional<UnsafeMutablePointer<UnsafeMutableRawPointer>>, but not the required Optional<UnsafeMutablePointer<Optional<UnsafeMutableRawPointer>>> (the difference is in that missing layer of "inner" nullability). To get around this, make your testHandle optional, yourself:

var testHandle: TESTHANDLE? // a.k.a. Optional<TESTHANDLE>, a.k.a. Optional< UnsafeMutableRawPointer>

然后，当您使用 & 运算符时，Swift 会将您的值包装在所需的 Optional<UnsafeMutablePointer<... >> 外层.

Then, when you use the & operator, Swift will wrap your value in the required Optional<UnsafeMutablePointer< ... >> outter layer.

typealias TESTHANDLE = UnsafeMutableRawPointer

func testInit(_ p: UnsafeMutablePointer<TESTHANDLE?>?) {
    print("Success!")
}

var testHandle: TESTHANDLE? = nil
testInit(&testHandle)

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