CODEWITHSUNDEEP
java
Bob
Bob

Reputation: 51

Java reading hex values into an array of type int

I have a file which contains integer numbers represented in hexadecimal IS there any way to store all of these numbers into an array of integers.

I know you can say int i = 0x

but I cannot do this when reading in the values I got an error?

Thanks in advance!

Upvotes: 5

Views: 12356

Answers (5)

Alex Cohn
Alex Cohn

Reputation: 57163

For strings that may have "0x" prefix, call Integer.decode(String). You can use it with Scanner,

try (Scanner s = new Scanner("0x11 0x22 0x33")) {
  while (s.hasNext()) {
    int num = Integer.decode(s.next());
    System.out.println(num);
  }
}
catch (Exception ex) {
  System.out.println(ex);
}

Unfortunately, unless the input is very short, Scanner is ridiculously slow. Here is an efficient hand-made parser for hexadecimal strings:

static boolean readArray(InputStream stream, int[] array) {

  int i = 0;

  final int SPACE = 0;
  final int X = 1;
  final int HEXNUM = 2;
  final int ERROR = -1;

  for (int next_char= -1, expected = SPACE; i <= array.length && stream.available() > 0 && expected != ERROR; next_char = stream.read()) {

    switch (expected)  {

      case SPACE:
        if (Character.isWhitespace(next_char))
          ;
        else if (next_char == '0') {
          array[i] = 0;
          expected = X;
        }
        else {
          LOGGER.e("unexpected '" + next_char + "' for " + expected + " at " + i);
          expected = ERROR;
        }
        break;

      case X:
        if (next_char == 'x' || next_char == 'X') {
          expected = HEXNUM;
        }
        else {
          LOGGER.e("unexpected '" + next_char + "' for " + expected + " at " + i);
          expected = ERROR;
        }
        break;

      case HEXNUM:
        if (Character.isDigit(next_char)) {
          array[i] *= 16;
          array[i] += next_char - '0';
        }
        else if (next_char >= 'a' && next_char <= 'f') {
          array[i] *= 16;
          array[i] += next_char - 'a' + 10;
        }
        else if (next_char >= 'A' && next_char <= 'F') {
          array[i] *= 16;
          array[i] += next_char - 'A' + 10;
        }
        else if (Character.isWhitespace(next_char)) {
          i++;
          expected = SPACE;
        }
        else {
          LOGGER.e("unexpected '" + next_char + "' for " + expected + " at " + i);
          expected = ERROR;
        }
    }
  }
}
if (expected == ERROR || i != array.length) {
  LOGGER.w("read " + i + " hexa integers when " + array.length + " were expected");
  return false;
}
return true;

Upvotes: 1

Starkey
Starkey

Reputation: 9781

Read in the values as strings, and call Integer.valueOf with a radix of 16.

See javadoc here: JavaSE6 Documentation: Integer.valueOf(String, int)

Upvotes: 0

Richard Fearn
Richard Fearn

Reputation: 25491

A Scanner might be useful. Are the numbers in your file prefixed with 0x? If so you'll have to remove that and convert to an integer:

// using StringReader for illustration; in reality you'd be reading from the file
String input = "0x11 0x22 0x33";
StringReader r = new StringReader(input);

Scanner s = new Scanner(r);
while (s.hasNext()) {
    String hexnum = s.next();
    int num = Integer.parseInt(hexnum.substring(2), 16);
    System.out.println(num);
}

If they're not prefixed with 0x it's even simpler:

String input = "11 22 33";
StringReader r = new StringReader(input);

Scanner s = new Scanner(r);
while (s.hasNext()) {
    int num = s.nextInt(16);
    System.out.println(num);
}

Upvotes: 0

aioobe
aioobe

Reputation: 421040

You probably want to go through Integer.parseInt(yourHexValue, 16).

Example:

// Your reader
BufferedReader sr = new BufferedReader(new StringReader("cafe\nBABE"));

// Fill your int-array
String hexString1 = sr.readLine();
String hexString2 = sr.readLine();

int[] intArray = new int[2];
intArray[0] = Integer.parseInt(hexString1, 16);
intArray[1] = Integer.parseInt(hexString2, 16);

// Print result (in dec and hex)
System.out.println(intArray[0] + " = " + Integer.toHexString(intArray[0]));
System.out.println(intArray[1] + " = " + Integer.toHexString(intArray[1]));

Output:

51966 = cafe
47806 = babe

Upvotes: 6

Bill K
Bill K

Reputation: 62769

I'm guessing you mean ascii hex? In that case there isn't a trivial way, but it's not hard.

You need to know exactly how the strings are stored in order to parse them.

IF they are like this:

1203 4058 a92e

then you need to read the file in and use spaces and linefeeds (whitespace) as separators.

If it's:

0x1203
0x4058

That's different yet

and if it's:

12034058...

That's something else.

Figure out how to get it into strings where each string ONLY contains the hex digits of a single number then call 

Integer.parseInt(string, 16)

Upvotes: 1

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