Binary to decimal converter

[Wayne Lam]

Hi guys.got some problem in this program.Was suppose to write a program to get a binary string and display its value. Need to have 3 methods.

binaryToDecimal()
-Receive string which represents the value
-Converts the binary value to decimal
-Return results

validateInput()
-Receive string as parameter
-Return true if string contains 0s and 1s,false otherwise

main()
-Read input using scanner class
-Call validateInput() method to check whether inputted value is binary value,display error message and repeat input process
-call binaryToDecimal() method to get its converted decimal value and display it


had some problem in the validateInput() part. not sure is it correct

[Wayne Lam]

import java.io.*;
import java.util.Scanner;

public class Q2
{
   public static void main(String[] args)
   {
	int power = 0;

	Scanner console = new Scanner(System.in);
    	
	while (true) {
	    System.out.println("Binary Converter");
	
	    System.out.print("Enter your Binary number : ");
	    String binaryInput = console.nextLine();	    
	
            if (binaryInput.equals("exit"))
                break;
    
    	    String strDigit = binaryInput.substring(binaryInput.length() - 1);
            int currentDigit = Integer.parseInt(strDigit); 
	
	    int tempPowerValue = 1;
	    for (int i = 1; i <= power; i++) {
		tempPowerValue *=  2;
	    }

	    System.out.println("Power value:" + tempPowerValue);	
            System.out.println("Last Digit: " + currentDigit);
            System.out.println("Decimal Value: " + binaryToDecimal(binaryInput));
        }
        System.out.println("Conversion program ended");
    }
       
    public static int binaryToDecimal(String binary) {
        return Integer.parseInt(binary, 2);
    }

    public static boolean validateInput(String input) {
	for(int i = 0; i < input.length(); i++) {
	
		char a = input.charAt(i) ;
		
		if (a == '0' || a == '1' ) {
			return true;
		}
	}
	
	// this will happen if all is 0s or 1s
	
	return false;
    }
}

[Ernie Mcracken]

for your validateInput method you could method something like this,

 public boolean validateString(String input){

   for (char c: input.toCharArray()){
       
        if (c!='0' && c!='1'){
             return false;
       }
   }

   return true;

}

Your current method will return every time you hit a 0 or 1 so it will only return false if the first character is not 1 or 0.

[Ulf Dittmer]

Or use Integer.valueOf(String, 16) and catch the exception.

I have edited your post to UseCodeTags; please do that in the future when posting code of an length. Otherwise it's unnecessarily hard to read the code, making it less likely that people will bother to do so.

[Ernie Mcracken]

Doe's he not need ;

Integer.parseInt(input, 2);

Given that he's trying to parse binary not hex??

[Ulf Dittmer]

Either parseInt or valueOf will work, but you're right of course that it needs to be 2, not 16.

[Wayne Lam]

Ernie Mcracken wrote:for your validateInput method you could method something like this,

 public boolean validateString(String input){

   for (char c: input.toCharArray()){
       
        if (c!='0' && c!='1'){
             return false;
       }
   }

   return true;

}

Your current method will return every time you hit a 0 or 1 so it will only return false if the first character is not 1 or 0.

Hi Ernie,after this,where should i call the method?

[Ernie Mcracken]

So you need to read your string using your Scanner then call the validateString() method

Scanner sc = new Scanner(System.in);
        String input;
        do{
            System.out.println("enter your string");
             input = sc.nextLine();

        }while (!validateString(input));

If you are going to have this in the same class as your Main then you'll need to make the validateString() method static.

[Wayne Lam]

Ernie Mcracken wrote:So you need to read your string using your Scanner then call the validateString() method

Scanner sc = new Scanner(System.in);
        String input;
        do{
            System.out.println("enter your string");
             input = sc.nextLine();

        }while (!validateString(input));

If you are going to have this in the same class as your Main then you'll need to make the validateString() method static.

Thanks Ernie again almost there,but now the problem is when i try to type "exit",the program doesn't exit.I need to display a message stating that "Please enter a proper binary number" when the validation returns false and allows the user to type another input.Appreciate the help as i quite new in programming.

import java.util.Scanner;
public class Q2
{
   public static void main(String[] args)
   {
	int power = 0;

	Scanner console = new Scanner(System.in);
    String binaryInput;
	
	
	while (true) {
	System.out.println("****Binary Converter****");
	
	do {
	
	System.out.print("Enter your Binary number : ");
	binaryInput = console.nextLine();	    
	
    } while (!validateInput(binaryInput)); 
    System.out.println("Please enter a valid binary number");
       
	
    if(binaryInput.equals("exit"))
    break;
        
	String strDigit = binaryInput.substring(binaryInput.length() - 1);
	int currentDigit = Integer.parseInt(strDigit); 	

	int tempPowerValue = 1;
	for(int i = 1; i <= power; i++){
    tempPowerValue *=  2;
	}

	System.out.println("Power value:" + tempPowerValue);	
    System.out.println("Last Digit: " + currentDigit);
    System.out.println("Decimal Value: " + binaryToDecimal(binaryInput));
    }
  System.out.println("Conversion program ended");
}
    
        
public static int binaryToDecimal(String binary){
return Integer.parseInt(binary, 2);

}

public static boolean validateInput(String input){  
  
    for (char c: input.toCharArray()){  
         
        if (c!='0' && c!='1'){  
             return false;  
        }  
        
    }  
    
    return true;  
   
 }  

[Ernie Mcracken]

Just put a line in your do while loop that says if the string equals exit then exit the program, you can do this with System.exit()

do {
	
	System.out.print("Enter your Binary number : ");
	binaryInput = console.nextLine();
	if (binaryInput.equals("exit") System.exit(0);    
	
    } while (!validateInput(binaryInput));