发出将Base64字符串转换为十六进制字符串的问题

Question

我已经设法使十六进制转换为十六进制，但您可以从输出中看到，当我尝试将Base64转换回十六进制时，有一个轻微的异常（例如，将Base64的最后四个值与十六进制输出进行比较）。

十六进制到基64:
应打印:SSdtIGtpbGxpbmcgeW91ciBicmFpbiBsaWtlIGEgcG9pc29ub3VzIG11c2hyb29t实际打印:SSDTIGTPBGXPBMCGEW91CIBICMFPBMCGEW91CIBICMFPBIBSAWTLIGEGCG9PC29UB3VZIG11C2HYB29T

Base64到十六进制:
应打印:49276D206B696C6C696E6720796F757220627261696E206C696B6520612070F6E6F7573206D757368726F6F6D
实际打印:49276D206B696C6C696E6720796E717220627261696E206C696B65206120706E69732E6E6F3573206C717328726F2E6D

我使用https://conv.darkbyte.ru/检查我的一些值，假设该站点上的代码是正确的，看来我的问题是如何从Base64获得Base10表示，而不是Base10到十六进制：

十进制等价

我的输出:
73,39,109,32,107,105,108,108,108,105,110,103,103,121,110,113,114,32,98,114,97,105,110,108,107,101,32,97,112,110,115,115,46,110,111,53,115,32,108,113,115,40,114,111,46,109

站点的输出:
73,39,109,32,107,105,108,108,108,105,110,103,32,121,111,117,114,32,98,114,97,105,110,32,108,107,101,32,112,111,111,111,111,111,111,111,111,111,111,117,115,117,115,117,117,115,104,114,111,111,109

似乎所有有错误的值都聚集在40-60和100-120附近，但我不确定从那里到底该往哪里走。我猜我正在处理某种边缘案件，但我不确定那会是什么。

相关代码：

    private static final Character[] base64Order = new Character[] { 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J',
        'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z', 'a', 'b', 'c', 'd', 'e',
        'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z',
        '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', '+', '/', };

    private static final Character[] hexOrder = new Character[] { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a',
        'b', 'c', 'd', 'e', 'f' };

public static String base64ToHex(String base64) throws Exception {
    if (base64.length() % 4 != 0 || base64.contains("[^a-zA-Z0-9\\+/]"))
        throw new Exception("InputNotBase64");
    else {
        int charValue = 0;
        int index = 0;
        String hex = "";
        BitSet bits = new BitSet();
        for (int i = 0; i < base64.length(); i++) {
            charValue = base64.charAt(i);
            // get actual value from ASCII table
            if (charValue > 64 && charValue < 91)
                charValue -= 65;
            if (charValue > 96 && charValue < 123)
                charValue -= 71;
            /// loop that adds to the BitSet reads right-to-left, so reverse
            // the bits and then shift
            charValue = Integer.reverse(charValue << 24) & 0xff;
            charValue >>= 2;
            // append binary values to the BitSet
            while (charValue != 0L) {
                if (charValue % 2 != 0) {
                    bits.set(index);
                }
                index++;
                charValue >>= 1;
            }
            // account for trailing 0s
            while (index % 6 != 0) {
                index++;
            }
        }
        // read 8-bit integer value for hex-value lookup
        String temp;
        int remainder;
        for (int i = 0; i < index; i++) {
            charValue = (charValue | (bits.get(i) ? 1 : 0));
            if ((i + 1) % 8 == 0) {
                temp = "";
                while (charValue != 0L) {
                    remainder = charValue % 16;
                    temp = hexOrder[remainder] + temp;
                    charValue /= 16;
                }
                hex += temp;
            }
            charValue <<= 1;
        }
        return hex;
    }
}

Answer 1

您忘记在代码中处理以下字符：“0”、“1”、“2”、“3”、“4”、“5”、“6”、“7”、“8”、“9”、“+”、“/”（如果替换以下代码

if (charValue > 64 && charValue < 91)
    charValue -= 65;
if (charValue > 96 && charValue < 123)
    charValue -= 71;

由

charValue = getPositionInBase64(charValue);

哪里

public static int getPositionInBase64(int n)
{
    for (int p = 0; p < base64Order.length; p++)
    {
        if (n == base64Order[p])
        {
            return p;
        }
    }
    return -1;
}

if (charValue >= 'A' && charValue <= 'Z')
    charValue -= 'A';
...

在这种情况下，发现问题更容易

因为你要求我提出可能的改进来加速计算。

准备下表并初始化一次

// index = character, value = index of character from base64Order
private static final int[] base64ToInt = new int[128];

public static void initBase64ToIntTable()
{
    for (int i = 0; i < base64Order.length; i++)
    {
        base64ToInt[base64Order[i]] = i;
    }
}

charValue = base64ToInt[base64.charAt(i)];

private static String intToHex(int n)
{
    return String.valueOf(new char[] { hexOrder[n/16], hexOrder[n%16] });
}

public static String base64ToHexVer2(String base64) throws Exception
{
    StringBuilder hex = new StringBuilder(base64.length()*3/4); //capacity could be 3/4 of base64 string length
    if (base64.length() % 4 != 0 || base64.contains("[^a-zA-Z0-9\\+/]"))
    {
        throw new Exception("InputNotBase64");
    }
    else
    {
        for (int i = 0; i < base64.length(); i += 4)
        {
            int n0 = base64ToInt[base64.charAt(i)];
            int n1 = base64ToInt[base64.charAt(i+1)];
            int n2 = base64ToInt[base64.charAt(i+2)];
            int n3 = base64ToInt[base64.charAt(i+3)];
            // in descriptions I treat all 64 base chars as 6 bit
            // all 6 bites from 0 and 1st 2 from 1st (00000011 ........ ........)
            hex.append(intToHex(n0*4 + n1/16));
            // last 4 bites from 1st and first 4 from 2nd (........ 11112222 ........)
            hex.append(intToHex((n1%16)*16 + n2/4));
            // last 2 bites from 2nd and all from 3rd (........ ........ 22333333)
            hex.append(intToHex((n2%4)*64 + n3));
        }
    }
    return hex.toString();
}

要测试速度，可以使用以下结构

    String b64 = "SSdtIGtpbGxpbmcgeW91ciBicmFpbiBsaWtlIGEgcG9pc29ub3VzIG11c2hyb29t";
    try
    {
        Base64ToHex.initBase64ToIntTable();
        System.out.println(Base64ToHex.base64ToHex(b64));
        System.out.println(Base64ToHex.base64ToHexVer2(b64));

        int howManyIterations = 100000;
        Date start, stop;
        long period;

        start = new Date();
        for (int i = 0; i < howManyIterations; i++)
        {
            Base64ToHex.base64ToHexVer2(b64);
        }
        stop = new Date();
        period = stop.getTime() - start.getTime();
        System.out.println("Ver2 taken " + period + " ms");

        start = new Date();
        for (int i = 0; i < howManyIterations; i++)
        {
            Base64ToHex.base64ToHex(b64);
        }
        stop = new Date();
        period = stop.getTime() - start.getTime();
        System.out.println("Ver1 taken " + period + " ms");

    }
    catch (Exception ex)
    {
    }

示例结果为

49276d206b696c6c696e6720796f757220627261696e206c696b65206120706f69736f6e6f7573206d757368726f6f6d
49276d206b696c6c696e6720796f757220627261696e206c696b65206120706f69736f6e6f7573206d757368726f6f6d
Ver2 taken 300 ms
Ver1 taken 2080 ms

但这只是近似值。当您首先检查Ver1,然后检查Ver2时,结果可能会有一点不同。此外，对于不同的java(6、7、8)和用于启动java的不同设置，结果也可能不同