iOS:在swift中将UnsafeMutablePointer 转换为String?
正如标题所说,在swift中将UnsafeMutablePointer转换为String的正确方法是什么?
//lets say x = UnsafeMutablePointer var str = x.memory.????
我尝试使用x.memory.description显然它是错误的,给我一个错误的字符串值。
如果指针指向以NUL结尾的UTF-8字节的C字符串,则可以执行以下操作:
import Foundation let x: UnsafeMutablePointer = ... // or UnsafePointer // or UnsafePointer // or UnsafeMutablePointer let str = String(cString: x)
时代变了。 在Swift 3+中你会这样做:
如果你想要validationutf-8:
let str: String? = String(validatingUTF8: c_str)
如果要将utf-8错误转换为unicode错误符号:
let str: String = String(cString: c_str)
假设c_str
的类型为UnsafePointer
或UnsafePointer
,它与大多数C函数返回的类型相同。
这个:
let str: String? = String(validatingUTF8: c_str)
似乎不适用于UnsafeMutablePointer
这是我简单地弄清楚如何做类似C / Perl系统function的事情:
let task = Process() task.launchPath = "/bin/ls" task.arguments = ["-lh"] let pipe = Pipe() task.standardOutput = pipe task.launch() let data = pipe.fileHandleForReading.readDataToEndOfFile() var unsafePointer = UnsafeMutablePointer.allocate(capacity: data.count) data.copyBytes(to: unsafePointer, count: data.count) let output : String = String(cString: unsafePointer) print(output) //let output : String? = String(validatingUTF8: unsafePointer) //print(output!)
如果我切换到validatingUTF8(可选)而不是cString,我收到此错误:
./ls.swift:19:37: error: cannot convert value of type 'UnsafeMutablePointer' to expected argument type 'UnsafePointer' (aka 'UnsafePointer') let output : String? = String(validatingUTF8: unsafePointer) ^~~~~~~~~~~~~
关于如何在管道输出上validationUTF8的想法(所以我没有在任何地方获得unicode错误符号)?
(是的,我没有正确检查我的打印选项(),这不是我目前正在解决的问题;-))。