Convert decimal to binary and vice-versa in Java

Question

I'm trying to learn Java, so I'm doing a few exercises from a programmer project idea book. This exercise requires that the program be able to convert a decimal number to binary and vice versa, like so:

Please enter the binary/decimal number to convert:9783569
100101010100100100010001
Please enter the binary/decimal number to convert:100101010100100100010001
9783569

Here's the code I came up with:

import java.util.Scanner;
public class BinaryToDecimalAndBackConverter{
 static String convertDecimalToBinary(String decimal){
 int integer = Integer.valueOf(decimal);
 String result = new String();
 while(integer > 0){
 result+=String.valueOf(integer%2);
 integer/=2;
 }
 result = new StringBuffer(result).reverse().toString();
 return result;
 }
 static int convertBinaryToDecimal(String binary){
 int result = 0;
 for(int reverseCounter = 0; reverseCounter < binary.length(); reverseCounter++){
 char currentChar = binary.charAt(binary.length() - reverseCounter - 1);
 int numericValue = Character.getNumericValue(currentChar);
 result+=Math.pow(2, reverseCounter) * numericValue;
// System.out.println(Math.pow(2, reverseCounter) * Character.getNumericValue(binary.charAt(binary.length() - reverseCounter - 1)) + " reverseCounter " + reverseCounter);
 }
 return result;
 }
 static String isDecimalOrBinary(String input){
 for(int counter = 0; counter < input.length(); counter++){
 if(input.charAt(counter) != '0' && input.charAt(counter) != '1'){
 return "decimal";
 }
 }
 return "binary";
 }
 public static void main(String[] args){
 Scanner inputScanner = new Scanner(System.in);
 System.out.print("Please enter the binary/decimal number to convert:");
 String input = inputScanner.nextLine();
 switch(isDecimalOrBinary(input)){
 case "decimal":
 System.out.println(convertDecimalToBinary(input));
 break;
 case "binary":
 System.out.println(convertBinaryToDecimal(input));
 break;
 }
 }
}

Answer 1

In addition to @Francesco Pitzalis's answer...

String concatenation

String result = new String();

This is never required as Strings are immutable in Java. If you find yourself using it to append String values, then you should be using StringBuilder (the faster, non-synchronized version of StringBuffer, which is usually recommended unless you require the multi-threading safety of the latter):

StringBuilder result = new StringBuilder();
// ...
result.append(integer % 2); // add some spaces for readability
// ...
return result.reverse().toString();

try-with-resources

// don't forget to close the Scanner
inputScanner.close();

Since you are on Java 7 at least (from the use of String in switch), you should be using try-with-resources to safely and efficiently manage the underlying I/O resource used by your Scanner instance:

try (Scanner scanner = new Scanner(System.in)) {
 // ...
}

Choices and enums

On a related note to your isDecimalOrBinary() method (which @Francesco Pitzalis's answer made a very good point), the return type can also be modeled as an enum should you still choose to keep it:

enum ValueType {
 DECIMAL, BINARY;
 public static ValueType parse(String value) {
 // ...
 }
}

This eliminates the possibility of typos in your expected String return values, e.g. when you return "decimal" but are wrongly checking for "Decimal" from calling the method.

• Sky: Code review never fails to teach me more about Java :). Thanks for your answer, especially for the try-with-resources part.

Answer 2

The best way to do it:

static String convertDecimalToBinary(String decimal) {
 return Integer.toString(Integer.parseInt(decimal, 10), 2);
}
static int convertBinaryToDecimal(String binary) {
 return Integer.parseInt(binary, 2);
}

By the way your code looks good except for at least two issues:

1. Don't work with negative numbers
2. The method isDecimalOrBinary is ambiguous: what's the result of the 100 input? It's binary of course, but if i meant one hundred? I suggest to delegate the user to specify it.

I would correct it in this way:

import java.util.Scanner;
public class BinaryToDecimalAndBackConverter {
 static String convertDecimalToBinary(String decimal) {
 final boolean isNegative = decimal.startsWith("-");
 if (isNegative) {
 decimal = decimal.substring(1);
 }
 int integer = Integer.valueOf(decimal);
 String result = new String();
 while (integer > 0) {
 result += String.valueOf(integer % 2);
 integer /= 2;
 }
 // use StringBuilder instead of StringBuffer, we are sure about
 // exclusive access, no need to use a thread safe class
 final StringBuilder resultBuilder = new StringBuilder(result);
 if (isNegative) {
 resultBuilder.append("-");
 }
 return resultBuilder.reverse().toString();
 }
 static int convertBinaryToDecimal(String binary) {
 final boolean isNegative = binary.startsWith("-");
 if (isNegative) {
 binary = binary.substring(1);
 }
 int result = 0;
 for (int reverseCounter = 0; reverseCounter < binary.length(); reverseCounter++) {
 final char currentChar = binary.charAt(binary.length() - reverseCounter - 1);
 final int numericValue = Character.getNumericValue(currentChar);
 result += Math.pow(2, reverseCounter) * numericValue;
 // System.out.println(Math.pow(2, reverseCounter) *
 // Character.getNumericValue(binary.charAt(binary.length() -
 // reverseCounter - 1)) + " reverseCounter " + reverseCounter);
 }
 if (isNegative) {
 result = -result;
 }
 return result;
 }
 public static void main(String[] args) {
 final Scanner inputScanner = new Scanner(System.in);
 System.out.print("Please enter the binary/decimal number to convert:");
 // expected input: NUM D|B
 final String[] splittedInput = inputScanner.nextLine().split(" ");
 if (splittedInput.length != 2) {
 // don't forget to close the Scanner
 inputScanner.close();
 throw new IllegalArgumentException("Wrong input format: expected NUM D|B");
 }
 final String input = splittedInput[0];
 switch (splittedInput[1]) {
 case "d":
 case "D":
 System.out.println(convertDecimalToBinary(input));
 break;
 case "b":
 case "B":
 System.out.println(convertBinaryToDecimal(input));
 break;
 default:
 // don't forget to close the Scanner
 inputScanner.close();
 throw new IllegalArgumentException(
 "Wrong input format: expected NUM D|B, the second argument was not D nor B");
 }
 // don't forget to close the Scanner
 inputScanner.close();
 }
}

• Sky: I didn't even consider the 100 case, so thanks for catching it and the unclosed scanner for me :). If only I had known how easy it is to convert the different number systems...I need to read more documentation. If no other answer come up, I'll accept this one :)