如何将字节数组转换为十六进制字符串,反之亦然?
当前回答
从微软的开发人员那里,一个很好的、简单的转换:
public static string ByteArrayToString(byte[] ba)
{
// Concatenate the bytes into one long string
return ba.Aggregate(new StringBuilder(32),
(sb, b) => sb.Append(b.ToString("X2"))
).ToString();
}
虽然上面的内容简洁紧凑,但性能狂热者会使用枚举器对此尖叫不已。通过Tomalak原始答案的改进版本,您可以获得最佳性能:
public static string ByteArrayToString(byte[] ba)
{
StringBuilder hex = new StringBuilder(ba.Length * 2);
for(int i=0; i < ba.Length; i++) // <-- Use for loop is faster than foreach
hex.Append(ba[i].ToString("X2")); // <-- ToString is faster than AppendFormat
return hex.ToString();
}
这是迄今为止我在这里看到的所有例程中速度最快的。不要只相信我的话…对每个例程进行性能测试并自行检查其CIL代码。
其他回答
我将参加这个比特拨弄比赛,因为我有一个同样使用比特拨弄来解码十六进制的答案。请注意,使用字符数组可能会更快,因为调用StringBuilder方法也需要时间。
public static String ToHex (byte[] data)
{
int dataLength = data.Length;
// pre-create the stringbuilder using the length of the data * 2, precisely enough
StringBuilder sb = new StringBuilder (dataLength * 2);
for (int i = 0; i < dataLength; i++) {
int b = data [i];
// check using calculation over bits to see if first tuple is a letter
// isLetter is zero if it is a digit, 1 if it is a letter
int isLetter = (b >> 7) & ((b >> 6) | (b >> 5)) & 1;
// calculate the code using a multiplication to make up the difference between
// a digit character and an alphanumerical character
int code = '0' + ((b >> 4) & 0xF) + isLetter * ('A' - '9' - 1);
// now append the result, after casting the code point to a character
sb.Append ((Char)code);
// do the same with the lower (less significant) tuple
isLetter = (b >> 3) & ((b >> 2) | (b >> 1)) & 1;
code = '0' + (b & 0xF) + isLetter * ('A' - '9' - 1);
sb.Append ((Char)code);
}
return sb.ToString ();
}
public static byte[] FromHex (String hex)
{
// pre-create the array
int resultLength = hex.Length / 2;
byte[] result = new byte[resultLength];
// set validity = 0 (0 = valid, anything else is not valid)
int validity = 0;
int c, isLetter, value, validDigitStruct, validDigit, validLetterStruct, validLetter;
for (int i = 0, hexOffset = 0; i < resultLength; i++, hexOffset += 2) {
c = hex [hexOffset];
// check using calculation over bits to see if first char is a letter
// isLetter is zero if it is a digit, 1 if it is a letter (upper & lowercase)
isLetter = (c >> 6) & 1;
// calculate the tuple value using a multiplication to make up the difference between
// a digit character and an alphanumerical character
// minus 1 for the fact that the letters are not zero based
value = ((c & 0xF) + isLetter * (-1 + 10)) << 4;
// check validity of all the other bits
validity |= c >> 7; // changed to >>, maybe not OK, use UInt?
validDigitStruct = (c & 0x30) ^ 0x30;
validDigit = ((c & 0x8) >> 3) * (c & 0x6);
validity |= (isLetter ^ 1) * (validDigitStruct | validDigit);
validLetterStruct = c & 0x18;
validLetter = (((c - 1) & 0x4) >> 2) * ((c - 1) & 0x2);
validity |= isLetter * (validLetterStruct | validLetter);
// do the same with the lower (less significant) tuple
c = hex [hexOffset + 1];
isLetter = (c >> 6) & 1;
value ^= (c & 0xF) + isLetter * (-1 + 10);
result [i] = (byte)value;
// check validity of all the other bits
validity |= c >> 7; // changed to >>, maybe not OK, use UInt?
validDigitStruct = (c & 0x30) ^ 0x30;
validDigit = ((c & 0x8) >> 3) * (c & 0x6);
validity |= (isLetter ^ 1) * (validDigitStruct | validDigit);
validLetterStruct = c & 0x18;
validLetter = (((c - 1) & 0x4) >> 2) * ((c - 1) & 0x2);
validity |= isLetter * (validLetterStruct | validLetter);
}
if (validity != 0) {
throw new ArgumentException ("Hexadecimal encoding incorrect for input " + hex);
}
return result;
}
从Java代码转换而来。
如果您希望比BitConverter更灵活,但不希望使用那些笨重的90年代风格的显式循环,那么您可以这样做:
String.Join(String.Empty, Array.ConvertAll(bytes, x => x.ToString("X2")));
或者,如果您使用的是.NET 4.0:
String.Concat(Array.ConvertAll(bytes, x => x.ToString("X2")));
(后者来自对原帖子的评论。)
我没有得到你建议的代码,Olipro。hex[i]+hex[i+1]显然返回了int。
然而,我确实从Waleeds代码中得到了一些提示,并将其结合在一起,取得了一些成功。这很难看,但根据我的测试(使用patricges测试机制),与其他测试相比,它似乎有1/3的时间在工作和执行。取决于输入大小。切换?:s首先将0-9分隔开可能会产生稍微更快的结果,因为数字比字母多。
public static byte[] StringToByteArray2(string hex)
{
byte[] bytes = new byte[hex.Length/2];
int bl = bytes.Length;
for (int i = 0; i < bl; ++i)
{
bytes[i] = (byte)((hex[2 * i] > 'F' ? hex[2 * i] - 0x57 : hex[2 * i] > '9' ? hex[2 * i] - 0x37 : hex[2 * i] - 0x30) << 4);
bytes[i] |= (byte)(hex[2 * i + 1] > 'F' ? hex[2 * i + 1] - 0x57 : hex[2 * i + 1] > '9' ? hex[2 * i + 1] - 0x37 : hex[2 * i + 1] - 0x30);
}
return bytes;
}
另一个快速功能。。。
private static readonly byte[] HexNibble = new byte[] {
0x0, 0x1, 0x2, 0x3, 0x4, 0x5, 0x6, 0x7,
0x8, 0x9, 0x0, 0x0, 0x0, 0x0, 0x0, 0x0,
0x0, 0xA, 0xB, 0xC, 0xD, 0xE, 0xF, 0x0,
0x0, 0x0, 0x0, 0x0, 0x0, 0x0, 0x0, 0x0,
0x0, 0x0, 0x0, 0x0, 0x0, 0x0, 0x0, 0x0,
0x0, 0x0, 0x0, 0x0, 0x0, 0x0, 0x0, 0x0,
0x0, 0xA, 0xB, 0xC, 0xD, 0xE, 0xF
};
public static byte[] HexStringToByteArray( string str )
{
int byteCount = str.Length >> 1;
byte[] result = new byte[byteCount + (str.Length & 1)];
for( int i = 0; i < byteCount; i++ )
result[i] = (byte) (HexNibble[str[i << 1] - 48] << 4 | HexNibble[str[(i << 1) + 1] - 48]);
if( (str.Length & 1) != 0 )
result[byteCount] = (byte) HexNibble[str[str.Length - 1] - 48];
return result;
}
扩展BigInteger方法(Gregory Morse在上面提到过)。我不能评论效率,它使用System.Linq.Reverse(),但它很小而且内置。
// To hex
byte[] bytes = System.Text.Encoding.UTF8.GetBytes("Test String!£");
string hexString = new System.Numerics.BigInteger(bytes.Reverse().ToArray()).ToString("x2");
// From hex
byte[] fromHexBytes = System.Numerics.BigInteger.Parse(hexString, System.Globalization.NumberStyles.HexNumber).ToByteArray().Reverse().ToArray();
// Unit test
CollectionAssert.AreEqual(bytes, fromHexBytes);
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