1. Write an application that inputs an integer between 00000 and 11111 and print

Question

1.      Write an application that inputs an integer between 00000 and 11111 and prints its decimal equivalent (between 0 and 63). If the number entered is out of range then the program must print an error message and allow the user to enter a new value.

Sample Run:

Enter a Binary Number between 00000 and 11111 (-1 to stop): 12343

WRONG VALUE! PLEASE ENTER A NUMBER AGAIN: 111111

WRONG VALUE! PLEASE ENTER A NUMBER AGAIN: 11000

The equivalent decimal number is: 24

Enter a Binary Number between 00000 and 11111 (-1 to stop): -1

Thank you !!!

Explanation / Answer

import java.util.*;

public class BinaryRep {

public static void main(String[] args) {

System.out.println("Enter a Binary Number between 00000 and 11111 (-1 to stop):");

while(true)

{

int flag=0;

Scanner input = new Scanner(System.in);

String number = input.nextLine();

String t="-1";

if((number.charAt(0)==t.charAt(0))&&(number.charAt(0)==t.charAt(0)))

break;

int len=number.length();

if(len!=5)

flag=1;

for(int i=0;i<5;i++)

if((number.charAt(i)!='1')&&(number.charAt(i)!='0'))

flag=1;

if(flag==0)

{int num=(int)((number.charAt(0)-'0')*16)+(int)((number.charAt(1)-'0')*8)+(int)((number.charAt(2)-'0')*4)+(int)((number.charAt(3)-'0')*2)+(int)(number.charAt(0)-'0');

System.out.println("the decimal equvalent is : "+num);

System.out.println("Enter a Binary Number between 00000 and 11111 (-1 to stop):");

}

else

{

System.out.println("WRONG VALUE! PLEASE ENTER A NUMBER AGAIN:");

}

}

}

}

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