Swift中的二进制到十六进制 [英] Binary to hexadecimal in Swift

问题描述

我有一个二进制字符串(例如"00100100")，我希望它以十六进制形式(如"24").

I have a string in binary (for example "00100100"), and I want it in hexadecimal (like "24").

是否编写了一种在Swift中将Binary转换为十六进制的方法?

Is there a method written to convert Binary to Hexadecimal in Swift?

推荐答案

可能的解决方案:

func binToHex(bin : String) -> String {
    // binary to integer:
    let num = bin.withCString { strtoul($0, nil, 2) }
    // integer to hex:
    let hex = String(num, radix: 16, uppercase: true) // (or false)
    return hex
}

只要数字适合UInt(32位或64位， 取决于平台).它使用BSD库函数 strtoul( )，它会根据给定的基数将字符串转换为整数.

This works as long as the numbers fit into the range of UInt (32-bit or 64-bit, depending on the platform). It uses the BSD library function strtoul() which converts a string to an integer according to a given base.

对于更大的数字，您必须处理输入 大块地.您可能还添加了输入字符串的 validation .

For larger numbers you have to process the input in chunks. You might also add a validation of the input string.

Swift 3/4的更新:不再需要strtoul功能. 返回nil表示无效输入:

Update for Swift 3/4: The strtoul function is no longer needed. Return nil for invalid input:

func binToHex(_ bin : String) -> String? {
    // binary to integer:
    guard let num = UInt64(bin, radix: 2) else { return nil }
    // integer to hex:
    let hex = String(num, radix: 16, uppercase: true) // (or false)
    return hex
}

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