我有一个使用哈希函数创建的字节数组。我想将此数组转换为字符串。到目前为止,这将给我十六进制字符串。
现在我想使用不同于十六进制字符的东西,我想使用这些36个字符对字节数组进行编码:[a-z][0-9]。
我该怎么做呢?
编辑:我之所以要这样做,是因为我想拥有比十六进制字符串更小的字符串。
我将我的任意长度进制转换函数从这个答案中移植到了C#:
static string BaseConvert(string number, int fromBase, int toBase)
{
var digits = "0123456789abcdefghijklmnopqrstuvwxyz";
var length = number.Length;
var result = string.Empty;
var nibbles = number.Select(c => digits.IndexOf(c)).ToList();
int newlen;
do {
var value = 0;
newlen = 0;
for (var i = 0; i < length; ++i) {
value = value * fromBase + nibbles[i];
if (value >= toBase) {
if (newlen == nibbles.Count) {
nibbles.Add(0);
}
nibbles[newlen++] = value / toBase;
value %= toBase;
}
else if (newlen > 0) {
if (newlen == nibbles.Count) {
nibbles.Add(0);
}
nibbles[newlen++] = 0;
}
}
length = newlen;
result = digits[value] + result; //
}
while (newlen != 0);
return result;
}
由于此代码是来自PHP,可能不太符合C#语言的惯用法,也没有参数有效性检查。但是,您可以将其提供给一个十六进制编码的字符串,它将能够正常工作。
var result = BaseConvert(hexEncoded, 16, 36);
虽然不完全是您所要求的,但将byte[]
编码为十六进制很简单。
var digits =“0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ”;
,为什么会得到相同的答案? - Kees C. Bakkerdigits
仅是可用于使用的数字。要实际使用大写字母,您需要将适当更高的基数作为第二个和/或第三个参数传递。 - Jonif (newlen == nibbles.Count) { nibbles.Add(0); }
这个判断语句?我看不出来为什么需要这样做。 - Payanewlen == nibbles.Count
并且您没有添加任何内容,那么紧接着访问 nibbles[newlen++]
就会抛出异常。 - Jonconst int kByteBitCount= 8; // number of bits in a byte
// constants that we use in FromBase36String and ToBase36String
const string kBase36Digits= "0123456789abcdefghijklmnopqrstuvwxyz";
static readonly double kBase36CharsLengthDivisor= Math.Log(kBase36Digits.Length, 2);
static readonly BigInteger kBigInt36= new BigInteger(36);
// assumes the input 'chars' is in big-endian ordering, MSB->LSB
static byte[] FromBase36String(string chars)
{
var bi= new BigInteger();
for (int x= 0; x < chars.Length; x++)
{
int i= kBase36Digits.IndexOf(chars[x]);
if (i < 0) return null; // invalid character
bi *= kBigInt36;
bi += i;
}
return bi.ToByteArray();
}
// characters returned are in big-endian ordering, MSB->LSB
static string ToBase36String(byte[] bytes)
{
// Estimate the result's length so we don't waste time realloc'ing
int result_length= (int)
Math.Ceiling(bytes.Length * kByteBitCount / kBase36CharsLengthDivisor);
// We use a List so we don't have to CopyTo a StringBuilder's characters
// to a char[], only to then Array.Reverse it later
var result= new System.Collections.Generic.List<char>(result_length);
var dividend= new BigInteger(bytes);
// IsZero's computation is less complex than evaluating "dividend > 0"
// which invokes BigInteger.CompareTo(BigInteger)
while (!dividend.IsZero)
{
BigInteger remainder;
dividend= BigInteger.DivRem(dividend, kBigInt36, out remainder);
int digit_index= Math.Abs((int)remainder);
result.Add(kBase36Digits[digit_index]);
}
// orientate the characters in big-endian ordering
result.Reverse();
// ToArray will also trim the excess chars used in length prediction
return new string(result.ToArray());
}
将“A test 1234. Made slightly larger!”编码为Base64结果为“165kkoorqxin775ct82ist5ysteekll7kaqlcnnu6mfe7ag7e63b5”。
在我的设备上,解码该Base64字符串1,000,000次需要12.6558909秒(我使用了与我的codereview答案提供的相同构建和机器条件)。
您提到您处理的是MD5哈希的byte[],而不是其十六进制字符串表示形式,因此我认为这个解决方案为您提供了最小的开销。
".IndexOf(chars[x])
,或者更好的是:使用一个映射到base36Digits
数组索引的每个base36字符代码的数组,可以提高FromBase36String
的性能。 - Dai{ 203, 77, 29, 30, 215, 37, 184, 136 }
编码为 "1taukyt738g1h"
,但是解码后变成了 { 53, 178, 226, 225, 40, 218, 71, 119 }
。 - Dai使用 BigInteger(需要 System.Numerics 引用)
const string chars = "0123456789abcdefghijklmnopqrstuvwxyz";
// The result is padded with chars[0] to make the string length
// (int)Math.Ceiling(bytes.Length * 8 / Math.Log(chars.Length, 2))
// (so that for any value [0...0]-[255...255] of bytes the resulting
// string will have same length)
public static string ToBaseN(byte[] bytes, string chars, bool littleEndian = true, int len = -1)
{
if (bytes.Length == 0 || len == 0)
{
return String.Empty;
}
// BigInteger saves in the last byte the sign. > 7F negative,
// <= 7F positive.
// If we have a "negative" number, we will prepend a 0 byte.
byte[] bytes2;
if (littleEndian)
{
if (bytes[bytes.Length - 1] <= 0x7F)
{
bytes2 = bytes;
}
else
{
// Note that Array.Resize doesn't modify the original array,
// but creates a copy and sets the passed reference to the
// new array
bytes2 = bytes;
Array.Resize(ref bytes2, bytes.Length + 1);
}
}
else
{
bytes2 = new byte[bytes[0] > 0x7F ? bytes.Length + 1 : bytes.Length];
// We copy and reverse the array
for (int i = bytes.Length - 1, j = 0; i >= 0; i--, j++)
{
bytes2[j] = bytes[i];
}
}
BigInteger bi = new BigInteger(bytes2);
// A little optimization. We will do many divisions based on
// chars.Length .
BigInteger length = chars.Length;
// We pre-calc the length of the string. We know the bits of
// "information" of a byte are 8. Using Log2 we calc the bits of
// information of our new base.
if (len == -1)
{
len = (int)Math.Ceiling(bytes.Length * 8 / Math.Log(chars.Length, 2));
}
// We will build our string on a char[]
var chs = new char[len];
int chsIndex = 0;
while (bi > 0)
{
BigInteger remainder;
bi = BigInteger.DivRem(bi, length, out remainder);
chs[littleEndian ? chsIndex : len - chsIndex - 1] = chars[(int)remainder];
chsIndex++;
if (chsIndex < 0)
{
if (bi > 0)
{
throw new OverflowException();
}
}
}
// We append the zeros that we skipped at the beginning
if (littleEndian)
{
while (chsIndex < len)
{
chs[chsIndex] = chars[0];
chsIndex++;
}
}
else
{
while (chsIndex < len)
{
chs[len - chsIndex - 1] = chars[0];
chsIndex++;
}
}
return new string(chs);
}
public static byte[] FromBaseN(string str, string chars, bool littleEndian = true, int len = -1)
{
if (str.Length == 0 || len == 0)
{
return new byte[0];
}
// This should be the maximum length of the byte[] array. It's
// the opposite of the one used in ToBaseN.
// Note that it can be passed as a parameter
if (len == -1)
{
len = (int)Math.Ceiling(str.Length * Math.Log(chars.Length, 2) / 8);
}
BigInteger bi = BigInteger.Zero;
BigInteger length2 = chars.Length;
BigInteger mult = BigInteger.One;
for (int j = 0; j < str.Length; j++)
{
int ix = chars.IndexOf(littleEndian ? str[j] : str[str.Length - j - 1]);
// We didn't find the character
if (ix == -1)
{
throw new ArgumentOutOfRangeException();
}
bi += ix * mult;
mult *= length2;
}
var bytes = bi.ToByteArray();
int len2 = bytes.Length;
// BigInteger adds a 0 byte for positive numbers that have the
// last byte > 0x7F
if (len2 >= 2 && bytes[len2 - 1] == 0)
{
len2--;
}
int len3 = Math.Min(len, len2);
byte[] bytes2;
if (littleEndian)
{
if (len == bytes.Length)
{
bytes2 = bytes;
}
else
{
bytes2 = new byte[len];
Array.Copy(bytes, bytes2, len3);
}
}
else
{
bytes2 = new byte[len];
for (int i = 0; i < len3; i++)
{
bytes2[len - i - 1] = bytes[i];
}
}
for (int i = len3; i < len2; i++)
{
if (bytes[i] != 0)
{
throw new OverflowException();
}
}
return bytes2;
}
BigInteger
一样重新创建临时变量。但仍然会很慢。问题在于编码第一个字节所需的时间是O(n),其中n是字节数组的长度(因为需要将整个数组除以36)。除非您想使用5个字节的块并且失去一些位。Base36的每个符号携带约5.169925001位。因此,8个这些符号将携带41.35940001位。非常接近40字节。0
(零)。请注意,如果您尝试使输出太小而无法容纳输入,则会抛出OverflowException
异常。+
和 =
)。 - C.EvenhuisSystem.Text.Encoding enc = System.Text.Encoding.ASCII;
string myString = enc.GetString(myByteArray);
你可以根据需要尝试不同的编码方式:
System.Text.ASCIIEncoding,
System.Text.UnicodeEncoding,
System.Text.UTF7Encoding,
System.Text.UTF8Encoding
为了满足要求 [a-z][0-9]
,您可以使用以下代码:
Byte[] bytes = new Byte[] { 200, 180, 34 };
string result = String.Join("a", bytes.Select(x => x.ToString()).ToArray());
您将拥有带有字符分隔符的字节字符串表示。要进行转换,您需要拆分并使用与 .Select()
相同的方法将 string[]
转换为 byte[]
。
250a244a...
。这只是一种选项,您可以用它来匹配[a-z][0-9]。 - Samich你可以使用模数。 这个例子将你的字节数组编码为[0-9][a-z]字符串。 如果需要,可以进行更改。
public string byteToString(byte[] byteArr)
{
int i;
char[] charArr = new char[byteArr.Length];
for (i = 0; i < byteArr.Length; i++)
{
int byt = byteArr[i] % 36; // 36=num of availible charachters
if (byt < 10)
{
charArr[i] = (char)(byt + 48); //if % result is a digit
}
else
{
charArr[i] = (char)(byt + 87); //if % result is a letter
}
}
return new String(charArr);
}
如果您不想因解码而丢失数据,可以使用此示例:
public string byteToString(byte[] byteArr)
{
int i;
char[] charArr = new char[byteArr.Length*2];
for (i = 0; i < byteArr.Length; i++)
{
charArr[2 * i] = (char)((int)byteArr[i] / 36+48);
int byt = byteArr[i] % 36; // 36=num of availible charachters
if (byt < 10)
{
charArr[2*i+1] = (char)(byt + 48); //if % result is a digit
}
else
{
charArr[2*i+1] = (char)(byt + 87); //if % result is a letter
}
}
return new String(charArr);
}
gResult = new Guid(oBytes);
,GUID 可以用尽可能少的32个十六进制字符表示sString = gResult.ToString("N");
。 - stevehipwell