16进制字符串：
00000018A0010098C68E00989A690000000000BC614E000055AA55AA
如何将上述16进制字符串转换成如下：
0x00 0x00 ...
HeapBuffer[pos=0 lim=28 cap=2048: 00 00 00 18 A0 01 00 98 C6 8E 00 98 9A 69 00 00...]
说明是要写入IoBuffer缓冲区。
求解决，在线等。。。。
------解决方案--------------------
- Java code
public class T { public static void main(String[] args) { String s = "00000018A0010098C68E00989A690000000000BC614E000055AA55AA"; System.out.println(s); byte[] b = HexString2Bytes(s); System.out.println(Bytes2HexString(b)); } /** * 将指定byte数组以16进制的形式打印到控制台 * * @param hint * String * @param b * byte[] * @return void */ public static void printHexString(String hint, byte[] b) { System.out.print(hint); for (int i = 0; i < b.length; i++) { String hex = Integer.toHexString(b[i] & 0xFF); if (hex.length() == 1) { hex = '0' + hex; } System.out.print(hex.toUpperCase() + " "); } System.out.println(""); } /** * * @param b * byte[] * @return String */ public static String Bytes2HexString(byte[] b) { String ret = ""; for (int i = 0; i < b.length; i++) { String hex = Integer.toHexString(b[i] & 0xFF); if (hex.length() == 1) { hex = '0' + hex; } ret += " 0x" + hex.toUpperCase(); } return ret; } /** * 将两个ASCII字符合成一个字节； 如："EF"–> 0xEF * * @param src0 * byte * @param src1 * byte * @return byte */ public static byte uniteBytes(byte src0, byte src1) { byte _b0 = Byte.decode("0x" + new String(new byte[] {src0})).byteValue(); _b0 = (byte) (_b0 << 4); byte _b1 = Byte.decode("0x" + new String(new byte[] { src1 })).byteValue(); byte ret = (byte) (_b0 ^ _b1); return ret; } /** * 将指定字符串src，以每两个字符分割转换为16进制形式 如："2B44EFD9" –> byte[]{0x2B, 0×44, 0xEF, * 0xD9} * * @param src * String * @return byte[] */ public static byte[] HexString2Bytes(String src) { if (null == src || 0 == src.length()) { return null; } byte[] ret = new byte[src.length() / 2]; byte[] tmp = src.getBytes(); for (int i = 0; i < (tmp.length / 2); i++) { ret[i] = uniteBytes(tmp[i * 2], tmp[i * 2 + 1]); } return ret; }}