Simple Java Issue

Question

This isn't working the way I expect it to. Sample input will include an asterisk that will resemble ether 0 or 1. The rest is comparing to see if the numbers match up. When I try to see if a character an asterisk it doesn't work the way I expect it to.

Samples:

org = *01

bit = 001 || 101 || 011

public static void doesItWork(String org, String bit)
    {
        for(int i = 0; i <= org.length() - 1; i++)
        {
            System.out.println(org.substring(i, i + 1));
            if(org.substring(i, i + 1) != "*" && org.substring(i, i + 1) !=  bit.substring(i, i + 1))
                break;

            if(i == org.length() - 1)
                System.out.println(bit);
        }
    }

Thanks guys for your help.

score 7 · Accepted Answer · edited May 23 '17 at 11:48

7

!= on Strings doesn't do what you want it to do. Java String.equals versus == explains, but in brief,

org.substring(i, i + 1) != "*"

should probably be

!org.substring(i, i + 1).equals("*")

or ideally

org.charAt(i) != '*'

edited May 23 '17 at 11:48

Community

1
1

answered Feb 10 '12 at 02:20

Mike Samuel

118,113
30
216
245

score 0 · Answer 2 · answered Feb 10 '12 at 02:21

0

!= with Strings might not be good idea. Try !x.equals("*");

answered Feb 10 '12 at 02:21

Ustaman Sangat

1,505
1
14
26

Also if you are a newcomer to java from a different language but know regular expressions, try them out in java, java.util.Regex – Ustaman Sangat Feb 10 '12 at 02:23

Simple Java Issue

2 Answers2