java – 问题将Base64字符串转换为十六进制字符串
作者：互联网
TLDR：
我遗漏的边缘情况是什么,或者我的算法中将Base64字符串转换为十六进制字符串是否有错误？
我最近决定尝试Matasano Crypto Challenges,但无论出于何种原因,我决定尝试编写第一个挑战,而不使用库来转换Hex和Base64字符串.
我已经设法让Hex到Base64转换工作,但是从输出中可以看出,当我尝试将Base64转换回Hex时会有轻微的异常(例如,将Base64的最后四个值与Hex输出进行比较) ).
Hex To Base64:
Should Print: SSdtIGtpbGxpbmcgeW91ciBicmFpbiBsaWtlIGEgcG9pc29ub3VzIG11c2hyb29t
Actually Prints: SSdtIGtpbGxpbmcgeW91ciBicmFpbiBsaWtlIGEgcG9pc29ub3VzIG11c2hyb29tBase64 to Hex:
Should Print: 49276d206b696c6c696e6720796f757220627261696e206c696b65206120706f69736f6e6f7573206d757368726f6f6d
Actually Prints: 49276d206b696c6c696e6720796e717220627261696e206c696b65206120706e69732e6e6f3573206c717328726f2e6d
我使用https://conv.darkbyte.ru/来检查我的一些值,并假设该站点上的代码是正确的,似乎我的问题是从Base64获取Base10表示,而不是从Base10到Hex：
Decimal Equivalent
My Output:
73, 39, 109, 32, 107, 105, 108, 108, 105, 110, 103, 32, 121, 110, 113, 114, 32, 98, 114, 97, 105, 110, 32, 108, 105, 107, 101, 32, 97, 32, 112, 110, 105, 115, 46, 110, 111, 53, 115, 32, 108, 113, 115, 40, 114, 111, 46, 109Site’s Output:
73, 39, 109, 32, 107, 105, 108, 108, 105, 110, 103, 32, 121, 111, 117, 114, 32, 98, 114, 97, 105, 110, 32, 108, 105, 107, 101, 32, 97, 32, 112, 111, 105, 115, 111, 110, 111, 117, 115, 32, 109, 117, 115, 104, 114, 111, 111, 109
似乎所有带错误的值都聚集在40-60和100-120之间,但我不确定从那里到底要去哪里.我猜是有一些边缘情况我会注意处理,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;
}
}
解决方法:
您忘记在代码中处理以下字符：’0′,’1′,’2′,’3′,’4′,’5′,’6′,’7′,’8′,’9′,’ ‘,’/’
如果您替换以下代码
if (charValue > 64 && charValue < 91)
charValue -= 65;
if (charValue > 96 && charValue < 123)
charValue -= 71;
通过
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;
}
一切正常
此外,当您使用字符而不是幻数时,代码更易读
if (charValue >= 'A' && charValue <= 'Z')
charValue -= 'A';
...
在这种情况下,发现问题更容易
因为你问我正在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;
}
}
现在您可以通过简单的操作替换您的if / else链
charValue = base64ToInt[base64.charAt(i)];
使用这个我写的方法比你的方法快几倍
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();
}
我认为这段代码更快,主要是因为简单的转换为十六进制.如果您想要并且需要它,您可以测试它.
要测试速度,您可以使用以下构造
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)
{
}
示例结果是
49276d206b696c6c696e6720796f757220627261696e206c696b65206120706f69736f6e6f7573206d757368726f6f6d
49276d206b696c6c696e6720796f757220627261696e206c696b65206120706f69736f6e6f7573206d757368726f6f6d
Ver2 taken 300 ms
Ver1 taken 2080 ms
但它只是近似值.当您首先检查Ver1而将Ver2检查为第二个时,结果可能会略有不同.另外,对于不同的javas(第6,7,8)以及用于启动java的不同设置,结果可能不同
标签：java,hex,base64,data-conversion 来源： https://codeday.me/bug/20190711/1432457.html