java arrayoutofbound exception after compiling

Question

package binarywa;
import java.util.*;

public class binaryadd {
    public static void main(String args[])
    {
        Scanner scan=new Scanner(System.in);
        String g=scan.next();
        String s[]=g.split("");
        int i,x=0;


        for(i=s.length;i>0;i--)
        {

         x+=Integer.parseInt(s[i])*Math.pow(2,i);
         }  

    }

}

Im trying to convert binary into decimal

but after compiling this im getting an arrayoutofbound exception

eg: 541656(its the input) Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 6 at binarywa.binaryadd.main(binaryadd.java:16)

`s.length` will always be outside the array. If `s = [1,2,3]`, `s.length = 3`, but `s[3]` doesnt exist. — Loocid, Apr 10 '18 at 05:20
@ScaryWombat i tried s.length-1 but then its not printing 1st digit — Abhishek Vishwakarma, Apr 10 '18 at 05:22
@AbhishekVishwakarma use this in your for loop. `for(i=s.length-1;i>=0;i--)` It will print all elements. See my answer for explanation — Sudoss, Apr 10 '18 at 05:35

score 1 · Accepted Answer · answered Apr 10 '18 at 05:24

keep the length s.length-1 and i>=0 will print the first element as array index start from 0

package binarywa;
import java.util.*;

public class binaryadd {
    public static void main(String args[])
    {
        Scanner scan=new Scanner(System.in);
        String g=scan.next();
        String s[]=g.split("");
        int i,x=0;


        for(i=s.length-1;i>=0;i--)
        {

         x+=Integer.parseInt(s[i])*Math.pow(2,i);
         }  

    }

}

score 0 · Answer 2 · answered Apr 10 '18 at 05:24

0

the for loop conditions are incorrect. Array indices are accessible from 0 to (length - 1). This is the reason you are getting ArrayIndexOutOfBoundsException

for (i = s.length - 1; i >= 0; i--)

answered Apr 10 '18 at 05:24

Onkar Kamatkar

202
2
10

score 0 · Answer 3 · answered Apr 10 '18 at 05:34

s.length in your code snipped will return 4.

In the loop you are trying to access s[4] which doesn't have a value. The s[] formed by g.split("") will have {1,2,3,4}

The indexes of these 4 array elements start with 0.

so,

s[0]=1
s[1]=2
s[2]=3
s[3]=4

once you try to access s[4] it will give ArrayIndexOutofBoundsException since at that index you don't have any value.

Solution is to run the loop as for(i=s.length-1;i>=0;i--)

score 0 · Answer 4 · answered Apr 10 '18 at 05:34

You need to change the for loop to as per the below

package binarywa;
import java.util.*;

public class binaryadd {
    public static void main(String args[])
    {
        Scanner scan=new Scanner(System.in);
        String g=scan.next();
        String s[]=g.split("");
        int i,x=0;


        for(i=s.length-1;i>=0;i--)
        {

         x+=Integer.parseInt(s[i])*Math.pow(2,i);
         }  

    }

}

But for that you also need to change your logic. i.e.

 x+=Integer.parseInt(s[i])*Math.pow(2,i+1);

else it will throw the NumberFormat Exception.

java arrayoutofbound exception after compiling

4 Answers4