Java字符串，单个字符到十六进制字节

Question

I want to convert a string by single char to 5 hex bytes and a byte represent a hex number:

like

String s = "ABOL1";

to

byte[] bytes = {41, 42, 4F, 4C, 01}

I tried following code , but Byte.decode got error when string is too large, like "4F" or "4C". Is there another way to convert it?

String s = "ABOL1";
char[] array = s.toCharArray();
for (int i = 0; i < array.length; i++) {
  String hex = String.format("%02X", (int) array[i]);
  bytes[i] = Byte.decode(hex);
}                

Answer 1

Is there any reason you are trying to go through string? Because you could just do this:

bytes[i] = (byte) array[i];

Or even replace all this code with:

byte[] bytes = s.getBytes(StandardCharsets.US_ASCII);

Answer 2

You can convert from char to hex String with String.format() :

String hex = String.format("%04x", (int) array[i]);

Or threat the char as an int and use:

String hex = Integer.toHexString((int) array[i]);

Answer 3

Use String hex = String.format("0x%02X", (int) array[i]); to specify HexDigits with 0x before the string.

A better solution is convert int into byte directly:

bytes[i] = (byte)array[i];

Answer 4

The Byte.decode() javadoc specifies that hex numbers should be on the form "0x4C" . So, to get rid of the exception, try this:

String hex = String.format("0x%02X", (int) array[i]);

There may also be a simpler way to make the conversion, because the String class has a method that converts a string to bytes :

bytes = s.getBytes();

Or, if you want a raw conversion to a byte array:

int i, len = s.length();
byte bytes[] = new byte[len];
String retval = name;
for (i = 0; i < len; i++) {
    bytes[i] = (byte) name.charAt(i);
}