快速块语法无法推断类型

Question

有人能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，唯一的区别是避免对UnsafePointer($0)的多次调用，而不是：

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)
}

带有错误信息：

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

正在调用的dn_function只是围绕来自libresolv的dn_expand的一个简单包装器。

public static func dn_expand(msg: UnsafePointer<UInt8>, eomorig: UnsafePointer<UInt8>, comp_dn: UnsafePointer<UInt8>, exp_dn: UnsafeMutablePointer<CChar>, length: Int) -> Int {
    return Int(res_9_dn_expand(msg, eomorig, comp_dn, exp_dn, Int32(length)))
}

Answer 1

正如注释中已经说过的，withUnsafePointer()不是获取指向元素存储的指针的正确方式。它编译，但给出意想不到的结果，如以下示例所示：

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

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

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

这将打印“随机”数字，但不输出预期的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, ...)
}

在这里，会自动推断闭包的返回类型，因为它是“单表达式”闭包。如果闭包包含多个表达式，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, ...)
}

或者让编译器从上下文中推断返回类型：

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