Thanks for your answer... I have 31 32 2E 30 31 33 6byte hexadecimal number. I want to convert 31 32 2E 30 31 33 this 6 byte hexadecimal number into 12.013 ASCII number in java.
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Why do you think that these numbers are equal? For the first look your first (hex) number is much much bigger then the second one.Roman– Roman2010-04-26 07:17:25 +00:00Commented Apr 26, 2010 at 7:17
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3 Answers
Something like this?
byte[] bytes = {0x31, 0x32, 0x2E, 0x30, 0x31, 0x33};
String result = new String(bytes, "ASCII");
System.out.println(result);
Comments
Probably not the most elegant method but try this:
char[6] string = new char[6];
string[0] = 0x31;
string[1] = 0x32;
string[2] = 0x2E;
string[3] = 0x30;
string[4] = 0x31;
string[5] = 0x33;
String s = new String(string);
int result = Integer.parseInt(s);
Comments
Assuming that your input is an array of strings representing the digits in hex, you can do the following:
public static String convert(String[] hexDigits){
byte[] bytes = new byte[hexDigits.length];
for(int i=0;i<bytes.length;i++)
bytes[i] = Integer.decode("0x"+hexDigits[i]).byteValue();
String result;
try {
result = new String(bytes, "ASCII");
} catch (UnsupportedEncodingException e) {
throw new RuntimeException(e);
}
return result;
}
Note that the code assumes that the digits are given as valid ASCII values, with no radix specifier.
