如何在Swift中将Int转换为十六进制字符串

In Obj-C I used to convert an unsigned integer n to a hex string with

在object - c中，我使用一个无符号整数n转换成十六进制字符串

 NSString *st = [NSString stringWithFormat:@"%2X", n];

I tried for a long time to translate this into Swift language, but unsuccessfully.

我花了很长时间把它翻译成快速的语言，但没有成功。

5 个解决方案

#1

You can now do:

你现在可以做的事:

let n = 14
var st = String(format:"%02X", n)
st += " is the hexadecimal representation of \(n)"
print(st)

0E is the hexadecimal representation of 14

Note: The 2 in this example is the field width and represents the minimum length desired. The 0 tells it to pad the result with leading 0's if necessary. (Without the 0, the result would be padded with leading spaces). Of course, if the result is larger than two characters, the field length will not be clipped to a width of 2; it will expand to whatever length is necessary to display the full result.

注意:本例中的2是字段宽度，表示所需的最小长度。如果需要，0告诉它以0开头的结果填充。(如果没有0，结果将被填充进前导空间)。当然，如果结果大于两个字符，则字段长度不会被裁剪为2的宽度;它将扩展到显示完整结果所需的任何长度。

This only works if you have Foundation imported (this includes the import of Cocoa or UIKit). This isn't a problem if you're doing iOS or macOS programming.

这只适用于导入Foundation(包括Cocoa或UIKit的导入)。如果你在做iOS或macOS编程，这不是问题。

Use uppercase X if you want A...F and lowercase x if you want a...f:

如果你想要一个…F和小写x，如果你想要…

String(format: "%x %X", 64206, 64206)  // "face FACE"

If you want to print integer values larger than UInt32.max, add ll (el-el, not eleven) to the format string:

如果您想打印大于UInt32的整数值。max，添加ll (el-el，不是11)到格式字符串:

let n = UInt64.max
print(String(format: "%llX is hexadecimal for \(n)", n))

FFFFFFFFFFFFFFFF is hexadecimal for 18446744073709551615

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Original Answer

原来的答案

You can still use NSString to do this. The format is:

你仍然可以使用NSString来做这个。格式是:

var st = NSString(format:"%2X", n)

This makes st an NSString, so then things like += do not work. If you want to be able to append to the string with += make st into a String like this:

这使得st成为一个NSString，因此诸如+=之类的东西不能工作。如果你想在字符串后面加上+= make st，就像这样:

var st = NSString(format:"%2X", n) as String

or

或

var st = String(NSString(format:"%2X", n))

or

或

var st: String = NSString(format:"%2X", n)

Then you can do:

然后你可以做:

let n = 123
var st = NSString(format:"%2X", n) as String
st += " is the hexadecimal representation of \(n)"
// "7B is the hexadecimal representation of 123"

#2

In Swift there is a specific init method on String for exactly this:

在Swift中，字符串上有一个特定的init方法:

let hex = String(0xF, radix: 16, uppercase: false)
println("hex=\(hex)") // Output: f

#3

With Swift 3, according to your needs, you may choose one of the three following methods in order to solve your problem.

使用Swift 3，根据您的需要，您可以选择以下三种方法之一来解决您的问题。

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#1. Using String's init(_:radix:uppercase:) initializer

Swift String has a init(_:radix:uppercase:) initializer with the following declaration:

Swift字符串有一个init(_:radix:upper - case:)初始化器，其声明如下:

init<T>(_ value: T, radix: Int = default, uppercase: Bool = default) where T : _SignedInteger

Creates a string representing the given value in base 10, or some other specified base.

创建一个字符串，表示以10为基数的给定值，或其他指定的基数。

The Playground code below shows how to create a String instance that represents an integer value in hexadecimal format by using init(_:radix:uppercase:) and without having to import Foundation:

下面的游戏代码演示了如何使用init(_:radix:大写:)来创建一个表示十六进制格式的整数值的字符串实例，而不需要导入基础:

let string1 = String(2, radix: 16)
print(string1) // prints: "2"

let string2 = String(211, radix: 16)
print(string2) // prints: "d3"

let string3 = String(211, radix: 16, uppercase: true)
print(string3) // prints: "D3"

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#2. Using String's init(format:​_:​) initializer

Foundation provides String a init(format:​_:​) initializer. init(format:​_:​) has the following declaration:

Foundation提供了一个init(格式:_:)初始化器字符串。init(格式:_:)具有以下声明:

init(format: String, _ arguments: CVarArg...)

Returns a String object initialized by using a given format string as a template into which the remaining argument values are substituted.

返回使用给定格式字符串作为模板初始化的字符串对象，其余参数值将被替换为模板。

The Apple's String Programming Guide gives a list of the format specifiers that are supported by String and NSString. Among those format specifiers, %X has the following description:

苹果的字符串编程指南给出了字符串和NSString支持的格式说明符的列表。在这些格式说明符中，%X有以下描述:

Unsigned 32-bit integer (unsigned int), printed in hexadecimal using the digits 0–9 and uppercase A–F.

无符号32位整数(无符号整数)，使用数字0-9和大写字母A-F以十六进制打印。

The Playground code below shows how to create a String instance that represents an integer value in hexadecimal format with init(format:​_:​):

下面的操场代码展示了如何创建一个字符串实例，它用init(格式:_:)以十六进制格式表示一个整数值:

import Foundation

let string1 = String(format:"%X", 2)
print(string1) // prints: "2"

let string2 = String(format:"%02X", 1)
print(string2) // prints: "01"

let string3 = String(format:"%02X", 211)
print(string3) // prints: "D3"

let string4 = String(format: "%02X, %02X, %02X", 12, 121, 255)
print(string4) // prints: "0C, 79, FF"

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#3. Using String's init(format:​arguments:​) initializer

Foundation provides String a init(format:​arguments:​) initializer. init(format:​arguments:​) has the following declaration:

Foundation提供了一个init(格式:parameters:)初始化器。init(格式:arguments:)具有以下声明:

init(format: String, arguments: [CVarArg])

Returns a String object initialized by using a given format string as a template into which the remaining argument values are substituted according to the user’s default locale.

返回使用给定格式字符串作为模板初始化的字符串对象，其余参数值将根据用户的默认语言环境替换为模板。

The Playground code below shows how to create a String instance that represents an integer value in hexadecimal format with init(format:​arguments:​):

下面的操场代码展示了如何创建一个字符串实例，它用init(格式:arguments:)以十六进制格式表示一个整数值:

import Foundation

let string1 = String(format:"%X", arguments: [2])
print(string1) // prints: "2"

let string2 = String(format:"%02X", arguments: [1])
print(string2) // prints: "01"

let string3 = String(format:"%02X",  arguments: [211])
print(string3) // prints: "D3"

let string4 = String(format: "%02X, %02X, %02X",  arguments: [12, 121, 255])
print(string4) // prints: "0C, 79, FF"

#4

To use

使用

let string2 = String(format:"%02X", 1)
print(string2) // prints: "01"

In Swift3 import foundation is not required, At least not in a Project. String should have all the functionality as NSString.

在Swift3 import foundation中不需要，至少在项目中不需要。String应该具有NSString的所有功能。

#5

Answers above work fine for values in the range of a 32 bit Int, but values over this won't work as the value will roll over.

上面的答案对于32位整型范围内的值是适用的，但是这个范围内的值不会起作用，因为值会滚动。

You need to use the length modifier for values greater than a 32bit Int

您需要使用大于32位整数的值的长度修改器。

%x = Unsigned 32-bit integer (unsigned int)

%x =无符号32位整数(无符号整数)

ll = Length modifiers specifying that a following d, o, u, x, or X conversion specifier applies to a long long or unsigned long long argument.

ll = Length修饰符，指定后面的d、o、u、x或x转换说明符应用于长长长或无符号长参数。

let hexString = String(format:"%llX", decimalValue)