Question

任何人都可以解释为什么这个带有隐式return的单行块编译：

let r = withUnsafePointer(&msg) {
    dn_expand(UnsafePointer($0), eomorig: UnsafePointer($0).advancedBy(msg.count), comp_dn: UnsafePointer($0).advancedBy(offset), exp_dn: &buf, length: buf.count)
}

但是这个版本重构了唯一的区别是避免多次调用UnsafePointer($0)不会：

let s = withUnsafePointer(&msg) {
    let p = UnsafePointer($0)
    return dn_expand(p, eomorig: p.advancedBy(msg.count), comp_dn: p.advancedBy(offset), exp_dn: &buf, length: buf.count)
}

错误消息：

Cannot convert value of type 'inout [UInt8]' (aka 'inout Array<UInt8>') to expected argument type 'inout _'

被调用的dn_function只是来自dn_expand的{{1}}的一个微不足道的包装：

libresolv

Answer 1

正如评论中已经说过的那样，withUnsafePointer() 不是获取指向元素存储的指针的正确方法。它编译，但是给出了意想不到的结果，如下例所示：

var msg: [UInt8] = [1, 2, 3, 4]

func foo(x: UnsafePointer<UInt8>) {
    print(x[0])
}

withUnsafePointer(&msg) {
    foo(UnsafePointer($0))
}

这打印＆＃34;随机＆＃34;数字，但不是预期的1。正确的方法是调用数组上的withUnsafeBufferPointer()方法：

msg.withUnsafeBufferPointer {
    foo($0.baseAddress)
}

在你的情况下

let r = msg.withUnsafeBufferPointer {
    dn_expand($0.baseAddress, eomorig: $0.baseAddress + $0.count, ...)
}

这里自动推断闭包的返回类型，因为它是一个单一的表达方式＆＃34;关闭。如果封闭包含更多如果不是一个表达式，则必须指定其类型：

let r = msg.withUnsafeBufferPointer { bufPtr -> Int in
    let p = bufPtr.baseAddress
    return dn_expand(p, eomorig: p + msg.count, ...)
}

或让编译器从上下文推断返回类型：

let r: Int = msg.withUnsafeBufferPointer { bufPtr in
    let p = bufPtr.baseAddress
    return dn_expand(p, eomorig: p + msg.count, ...)
}

Swift块语法无法推断类型

1 个答案: