[英]Convert java string with odd length to hex byte array
I need to convert string to hexademical byte array,my code is: 我需要将字符串转换为十六进制字节数组,我的代码是:
public static byte[] stringToHex(final String buf)
{
return DatatypeConverter.parseHexBinary(buf);
}
According to java doc to convert string to Hex DatatypeConverter
use the following implementation 根据java doc将字符串转换为Hex
DatatypeConverter
使用以下实现
public byte[] parseHexBinary(String s) {
final int len = s.length();
// "111" is not a valid hex encoding.
if (len % 2 != 0) {
throw new IllegalArgumentException("hexBinary needs to be even-length: " + s);
}
byte[] out = new byte[len / 2];
for (int i = 0; i < len; i += 2) {
int h = hexToBin(s.charAt(i));
int l = hexToBin(s.charAt(i + 1));
if (h == -1 || l == -1) {
throw new IllegalArgumentException("contains illegal character for hexBinary: " + s);
}
out[i / 2] = (byte) (h * 16 + l);
}
return out;
}
It means that only strings with the even length is legal to be converted.But in php there is no such constraint For example code in php: 这意味着只有偶数长度的字符串才可以合法转换。但是在php中没有这样的约束。例如php中的代码:
echo pack("H*", "250922f67dcbc2b97184464a91e7f8f");
And in java 在java中
String hex = "250922f67dcbc2b97184464a91e7f8f";
System.out.println(stringToHex(hex));//my method that was described earlier
Why the following string is legal in php? 为什么以下字符串在php中合法?
PHP just adds a final 0
in case the number of characters is odd. 如果字符数为奇数,PHP只会添加最后一个
0
。
Both of these 这两个
echo pack("H*", "48454C50");
echo pack("H*", "48454C5");
yield 让
HELP
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