macos 如何将 NSData 转换为 NSString Hex 字符串？
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How to convert an NSData into an NSString Hex string?
提问by Todd Ditchendorf
When I call -descriptionon an NSDataobject, I see a pretty Hex string of the NSDataobject's bytes like:
当我调用-description一个NSData对象时，我看到一个漂亮的NSData对象字节的十六进制字符串，例如：
<f6e7cd28 0fc5b5d4 88f8394b af216506 bc1bba86 4d5b483d>
I'd like to get this representation of the data (minus the lt/gt quotes) into an in-memory NSStringso I can work with it.. I'd prefer not to call -[NSData description]and then just trim the lt/gt quotes (because I assume that is not a guaranteed aspect of NSData's public interface and is subject change in the future).
我想将数据的这种表示（减去 lt/gt 引号）放入内存中，NSString以便我可以使用它..我不想调用-[NSData description]然后只修剪 lt/gt 引号（因为我认为这不是NSData的公共接口的保证方面，并且将来会发生变化）。
What's the simplest way to get this representation of an NSDataobject into an NSStringobject (other than calling -description)?
将NSData对象的这种表示形式转换为NSString对象的最简单方法是什么（调用-description除外）？
回答by Erik Aigner
Keep in mind that any String(format: ...)solution will be terribly slow (for large data)
请记住，任何String(format: ...)解决方案都会非常慢（对于大数据）
NSData *data = ...;
NSUInteger capacity = data.length * 2;
NSMutableString *sbuf = [NSMutableString stringWithCapacity:capacity];
const unsigned char *buf = data.bytes;
NSInteger i;
for (i=0; i<data.length; ++i) {
[sbuf appendFormat:@"%02X", (NSUInteger)buf[i]];
}
If you need something more performanttry this:
如果您需要更高性能的东西，请尝试以下操作：
static inline char itoh(int i) {
if (i > 9) return 'A' + (i - 10);
return '0' + i;
}
NSString * NSDataToHex(NSData *data) {
NSUInteger i, len;
unsigned char *buf, *bytes;
len = data.length;
bytes = (unsigned char*)data.bytes;
buf = malloc(len*2);
for (i=0; i<len; i++) {
buf[i*2] = itoh((bytes[i] >> 4) & 0xF);
buf[i*2+1] = itoh(bytes[i] & 0xF);
}
return [[NSString alloc] initWithBytesNoCopy:buf
length:len*2
encoding:NSASCIIStringEncoding
freeWhenDone:YES];
}
Swift 4.2 version
斯威夫特 4.2 版本
extension Data {
var hexString: String? {
return withUnsafeBytes { (bytes: UnsafePointer<UInt8>) in
let charA = UInt8(UnicodeScalar("a").value)
let char0 = UInt8(UnicodeScalar("0").value)
func itoh(_ value: UInt8) -> UInt8 {
return (value > 9) ? (charA + value - 10) : (char0 + value)
}
let hexLen = count * 2
let ptr = UnsafeMutablePointer<UInt8>.allocate(capacity: hexLen)
for i in 0 ..< count {
ptr[i*2] = itoh((bytes[i] >> 4) & 0xF)
ptr[i*2+1] = itoh(bytes[i] & 0xF)
}
return String(bytesNoCopy: ptr,
length: hexLen,
encoding: .utf8,
freeWhenDone: true)
}
}
}
回答by AliSoftware
I agree on the solution notto call descriptionwhich is to be reserved for debugging, so good point and good question :)
我同意不调用的解决方案，description这是为调试保留的，所以好点和好问题:)
The easiest solution is to loop thru the bytes of the NSDataand construct the NSString from it. Use [yourData bytes]to access the bytes, and build the string into an NSMutableString.
最简单的解决方案是循环遍历 的字节NSData并从中构造 NSString。使用[yourData bytes]访问字节，并建立串入一个NSMutableString。
Here is an example by implementing this using a category of NSData
这是一个使用 NSData 类别实现的示例
@interface NSData(Hex)
-(NSString*)hexRepresentationWithSpaces_AS:(BOOL)spaces;
@end
@implementation NSData(Hex)
-(NSString*)hexRepresentationWithSpaces_AS:(BOOL)spaces
{
const unsigned char* bytes = (const unsigned char*)[self bytes];
NSUInteger nbBytes = [self length];
//If spaces is true, insert a space every this many input bytes (twice this many output characters).
static const NSUInteger spaceEveryThisManyBytes = 4UL;
//If spaces is true, insert a line-break instead of a space every this many spaces.
static const NSUInteger lineBreakEveryThisManySpaces = 4UL;
const NSUInteger lineBreakEveryThisManyBytes = spaceEveryThisManyBytes * lineBreakEveryThisManySpaces;
NSUInteger strLen = 2*nbBytes + (spaces ? nbBytes/spaceEveryThisManyBytes : 0);
NSMutableString* hex = [[NSMutableString alloc] initWithCapacity:strLen];
for(NSUInteger i=0; i<nbBytes; ) {
[hex appendFormat:@"%02X", bytes[i]];
//We need to increment here so that the every-n-bytes computations are right.
++i;
if (spaces) {
if (i % lineBreakEveryThisManyBytes == 0) [hex appendString:@"\n"];
else if (i % spaceEveryThisManyBytes == 0) [hex appendString:@" "];
}
}
return [hex autorelease];
}
@end
Usage:
用法：
NSData* data = ...
NSString* hex = [data hexRepresentationWithSpaces_AS:YES];
回答by Johannes Lund
Just wanted to add that @PassKits's method can be written very elegantly using Swift 3 since Datanow is a collection.
只是想补充一点，@PassKits 的方法可以使用 Swift 3 非常优雅地编写，因为Data现在是一个集合。
extension Data {
var hex: String {
var hexString = ""
for byte in self {
hexString += String(format: "%02X", byte)
}
return hexString
}
}
Or ...
或者 ...
extension Data {
var hex: String {
return self.map { b in String(format: "%02X", b) }.joined()
}
}
Or even ...
甚至 ...
extension Data {
var hex: String {
return self.reduce("") { string, byte in
string + String(format: "%02X", byte)
}
}
}
回答by ma11hew28
I liked @Erik_Aigner's answer the best. I just refactored it a bit:
我最喜欢@Erik_Aigner 的回答。我只是重构了一下：
NSData *data = [NSMutableData dataWithBytes:"acani" length:5];
NSUInteger dataLength = [data length];
NSMutableString *string = [NSMutableString stringWithCapacity:dataLength*2];
const unsigned char *dataBytes = [data bytes];
for (NSInteger idx = 0; idx < dataLength; ++idx) {
[string appendFormat:@"%02x", dataBytes[idx]];
}
回答by PassKit
In Swift you can create an extension.
在 Swift 中，您可以创建扩展。
extension NSData {
func toHexString() -> String {
var hexString: String = ""
let dataBytes = UnsafePointer<CUnsignedChar>(self.bytes)
for (var i: Int=0; i<self.length; ++i) {
hexString += String(format: "%02X", dataBytes[i])
}
return hexString
}
}
Then you can simply use:
然后你可以简单地使用：
let keyData: NSData = NSData(bytes: [0x00, 0xFF], length: 2)
let hexString = keyData.toHexString()
println("\(hexString)") // Outputs 00FF
回答by Lily Ballard
Sadly there's no built-in way to produce hex from an NSData, but it's pretty easy to do yourself. The simple way is to just pass successive bytes into sprintf("%02x") and accumulate those into an NSMutableString. A faster way would be to build a lookup table that maps 4 bits into a hex character, and then pass successive nybbles into that table.
遗憾的是，没有内置的方法可以从 NSData 生成十六进制，但是自己很容易做到。简单的方法是将连续字节传递给 sprintf("%02x") 并将它们累积到 NSMutableString 中。一种更快的方法是构建一个查找表，将 4 位映射到一个十六进制字符，然后将连续的 nybbles 传递到该表中。
回答by Steve Streza
While it may not be the most efficient way to do it, if you're doing this for debugging, SSCrypto has a category on NSData which contains two methods to do this (one for creating an NSString of the raw byte values, and one which shows a prettier representation of it).
虽然这可能不是最有效的方法，但如果您这样做是为了调试，SSCrypto 在 NSData 上有一个类别，其中包含两种方法（一种用于创建原始字节值的 NSString，另一种用于显示了它的更漂亮的表示）。
回答by pheuberger
Seeing there is a Swift 1.2 snippet in the comments, here's the Swift 2 version since C style for loops are deprecated now. Gistwith MIT license and two simple unit tests if you care.
看到评论中有一个 Swift 1.2 片段，这里是 Swift 2 版本，因为现在不推荐使用 C 风格的 for 循环。 吉斯特与MIT许可证，如果你关心两个简单的单元测试。
Here's the code for your convenience:
为了您的方便，这是代码：
import Foundation
extension NSData {
var hexString: String {
let pointer = UnsafePointer<UInt8>(bytes)
let array = getByteArray(pointer)
return array.reduce("") { (result, byte) -> String in
result.stringByAppendingString(String(format: "%02x", byte))
}
}
private func getByteArray(pointer: UnsafePointer<UInt8>) -> [UInt8] {
let buffer = UnsafeBufferPointer<UInt8>(start: pointer, count: length)
return [UInt8](buffer)
}
}
