将奇数长度的java字符串转换为十六进制字节数组

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 DatatypeConverteruse the following implementation

根据java doc将字符串转换为Hex DatatypeConverteruse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:

这意味着只有具有偶数长度的字符串才能被转换。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中是合法的?

1 个解决方案

#1

PHP just adds a final 0 in case the number of characters is odd.

如果字符数为奇数,PHP只会添加最后的0。

Both of these

这两个

echo pack("H*", "48454C50");
echo pack("H*", "48454C5");

yield

HELP

#1