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Java Exercises: Convert a binary number to a Octal number

Java Basic: Exercise-24 with Solution

Write a Java program to convert a binary number to a Octal number.

Binary number: A binary number is a number expressed in the base-2 numeral system or binary numeral system. This system uses only two symbols: typically 0(zero) and 1(one).

Octal number: The octal numeral system is the base-8 number system, and uses the digits 0 to 7.

Test Data:
Input a binary number: 111

Pictorial Presentation: Binary to Octal number

Java: Convert a binary number to a Octal number

Sample Solution:

Java Code:

import java.util.*;
public class Exercise24 {
public static void main(String[] args) 
    {
        int binnum, binnum1,rem, decnum=0, quot, i=1, j;
        int octnum[] = new int[100];
        Scanner scan = new Scanner(System.in);
		System.out.print("Input a Binary Number : ");
        binnum = scan.nextInt();
        binnum1=binnum;
     
      while(binnum > 0)
        {
            rem = binnum % 10;
            decnum = decnum + rem*i;
            //System.out.println(rem);
            i = i*2;
            binnum = binnum/10;
        }   

		i=1;
        quot = decnum;
		
        while(quot > 0)
        {
            octnum[i++] = quot % 8;
            quot = quot / 8;
        }
		
        System.out.print("Equivalent Octal Value of " +binnum1+ " is :");
        for(j=i-1; j>0; j--)
        {
            System.out.print(octnum[j]);
        }
 System.out.print("\n");
 
    }
}

Sample Output:

Enter Binary Number : 111                                                                                     
Equivalent Octal Value of 111 is :7  

Flowchart:

Flowchart: Java exercises: Convert a binary number to a Octal number

Java Code Editor:

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Previous: Write a Java program to convert a binary number to hexadecimal number.
Next: Write a Java program to convert a octal number to a decimal number.

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Java: Tips of the Day

Try and catch:

Java is excellent at catching errors, but it can only recover gracefully if you tell it what to do. The cascading hierarchy of attempting to perform an action in Java starts with try, falls back to catch, and ends with finally. Should the try clause fail, then catch is invoked, and in the end, there's always finally to perform some sensible action regardless of the results. Here's an example:

try {
        cmd = parser.parse(opt, args); 
       
        if(cmd.hasOption("help")) {
                HelpFormatter helper = new HelpFormatter();
                helper.printHelp("Hello ", opt);
                System.exit(0);
                }
        else {
                if(cmd.hasOption("shell") || cmd.hasOption("s")) {
                String target = cmd.getOptionValue("tgt");
                } // else
        } // fi
} catch (ParseException err) {
        System.out.println(err);
        System.exit(1);
        } //catch
        finally {
                new Hello().helloWorld(opt);
        } //finally
} //try

It's a robust system that attempts to avoid irrecoverable errors or, at least, to provide you with the option to give useful feedback to the user. Use it often, and your users will thank you!

Ref: https://red.ht/3EZc9OC