正如标题所说,Swift中将UnsafeMutablePointer转换为String的正确方法是什么?
//lets say x = UnsafeMutablePointer<Int8>
var str = x.memory.????
我尝试使用x.memory.description,但很明显是错误的,给了我一个错误的字符串值。
如果指针指向以UTF-8字节为结尾的C字符串,您可以这样做:
import Foundation
let x: UnsafeMutablePointer<Int8> = ...
// or UnsafePointer<Int8>
// or UnsafePointer<UInt8>
// or UnsafeMutablePointer<UInt8>
let str = String(cString: x)
时代在变化。在Swift 3+中,您需要这样做:
如果您想要验证utf-8:
let str: String? = String(validatingUTF8: c_str)
如果你想将utf-8错误转化为Unicode错误符号:�
let str: String = String(cString: c_str)
假设 c_str 的类型为 UnsafePointer<UInt8>
或 UnsafePointer<CChar>,这是相同的类型,并且大多数C函数返回。
let str: String? = String(validatingUTF8: c_str)
似乎不能与UnsafeMutablePointer<UInt8>一起使用(这是我的数据中出现的内容)。
这让我轻松地想到如何执行类似于C/Perl系统函数的操作:
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<Int8>.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!)
./ls.swift:19:37: error: cannot convert value of type 'UnsafeMutablePointer<UInt8>' to expected argument type 'UnsafePointer<CChar>' (aka 'UnsafePointer<Int8>')
let output : String? = String(validatingUTF8: unsafePointer)
^~~~~~~~~~~~~
如何验证管道输出的UTF8格式(以便我不会在任何地方看到Unicode错误符号)?
(是的,我没有正确检查print()的可选项,但这不是我目前解决的问题;-)。)
String(CString:, encoding:)
。 - Rob