Converting char array to int

[Roger F. Gay]

I've been staring at this for hours, trying this and that, and still don't see what's wrong. I know that it's ok to:

int newvalue = Integer.parseInt(new String(char[] something));

But in the context of this code, I get a NumberFormatException. As you can see, I've been looking at the character array and String and don't see any problem with it. When I simplify (beyond what is possible in the actual application) by just setting a string value to the number that I get from the algorithm, all is fine. When I create a char[] array and set it, all is fine. Just can't see why this is throwing the exception.

import java.io.*;
import java.net.*;

public class CodeTest {

  public static void main (String[] args) throws IOException {

    String key = "38w- 8 dN4 25 26:8e J  0";

    int num_spaces=0;
    int j=9;
    int newvalue;
    char[] newnumber = new char[10];
    for (int i=key.length()-1; i>=0; i--) {
      if (Character.isSpaceChar(key.charAt(i))) {
        ++num_spaces;
      } else if (Character.isDigit(key.charAt(i))) {
        newnumber[j--] = key.charAt(i);
      }
    }
    for (int i=0; i<10; i++) {
      System.out.print(newnumber[i]);
    }
    System.out.println(" ");

    String thisun = new String(newnumber);
     System.out.println("thisun = " + thisun);

    try {
    
      newvalue = Integer.parseInt(thisun)/num_spaces;

      System.out.println("Successful run, newvalue = " + newvalue);

    } catch (NumberFormatException e) {
        System.out.println("NumberFormatException:WSHandshakeHandler:keyCrypter32: " + e.getMessage());
    }
  }
}

[Ogeh Ikem]

The number (3884252680) looks too big for a Java Integer.

[Unnar Björnsson]

That's because you are trying to convert the string "3884252680" to an int where the largest value int can hold is 2147483647

[Roger F. Gay]

DO-OH! I knew it had to be something basic. All I had in mind was that the number had to hold 10 decimal places or less. Thanks.

What can I do to fix it? I don't think there's a BigInteger.parseBigInt().

[Rob Spoor]

Long.parseLong?

[Roger F. Gay]

I fixed the posted problem by using long instead of int. But I'm concerned (and don't know a big-endian from a little indian). The test process came from a real example. From the spec:

For each of these fields, the server has to take the digits from the
value to obtain a number (in this case 1868545188 and 1733470270
respectively), then divide that number by the number of spaces
characters in the value (in this case 12 and 10) to obtain a 32-bit
number (155712099 and 173347027). ,,,,,

.....

The concatenation of the number obtained from processing the |Key1| field,
expressed as a big-endian 32-bit number, the
number obtained from processing the |Key2| field, again
expressed as a big-endian 32-bit number, and finally the eight bytes
at the end of the handshake, form a 128 bit string .....

[Roger F. Gay]

Never mind, I guess. After dividing by the number of spaces, the result is 32-bit. No problem there. I just panicked because I don't really understand what a big-endian is and whether I have to do anything special.

[Unnar Björnsson]

Roger F. Gay wrote:Never mind, I guess. After dividing by the number of spaces, the result is 32-bit. No problem there. I just panicked because I don't really understand what a big-endian is and whether I have to do anything special.

Big endian logic stores the most significant byte (MSB) first. An int number is 32bit which means it takes 4 bytes in memory, so the number 123456789 (75BCD15 in hex) would be stored as [07] [5B] [CD] [15] where little-endian would store it as [15] [CD] [5B] [07]