Binary Converter Java

Question

So i have a problem with finishing this project this is how I did my code:
import java.util.Scanner;
//your code goes here
public class Converter {
public static toBinary(int num) {
String binary="" ;
while(num > 0){
binary = (num%2)+binary;
num/=2;
}
return binary;
}
}
public class Program{
public static void main(String[ ] args) {
Scanner sc= new Scanner(System.in);
int x=scr.nextIn();
System.out.println (Converter.toBinary(x));
}
}
But I don't get any output with this code! Would appreciate any advice, thank you in advance!

Answer 1

You have some typos and return type is missing.
Return type is String and the typos i spotted are on reading input so the line should be:
int x = sc.nextInt();

Answer 2

import java.util.Scanner;

public class Converter {
public static String toBinary(int num) {
String binary="";
while(num > 0) {
binary = (num%2)+binary;
num /= 2;
}
return binary;
}
}

public class Program {
public static void main(String[ ] args) {
Scanner sc = new Scanner(System.in);
int x = sc.nextInt();
System.out.print(Converter.toBinary(x));
}
}

Answer 3

import java.util.Scanner;

//your code goes here
class Converter{
static String toBinary(int x){
String bin="";
while(x>0){
bin=(x%2)+bin;
x/=2;
}
return bin;
}
}

public class Program {
public static void main(String[ ] args) {
Scanner sc = new Scanner(System.in);
int x = sc.nextInt();
System.out.print(Converter.toBinary(x));
}
}

Answer 4

Hi everybody. Can anyone explain, why the type of returned value in the toBinary method is String? Shouldn't it return integer? Since we are working with integers? Or am I missing something?

Answer 5

Ok. I've found kind of answer here:
https://www.sololearn.com/Discuss/2923028/?ref=app
But I still don't really get, why we use String to avoid exceptions with long numbers. String is, well... String. Doesn't have anything to do with numbers. ?

Edit.
Ok. After a long thinking, I think I understand. It's more memory efficient than using a return type long. I suppose.

Answer 6

We use string to make the binary digits visible. In memory everything is binary, even the characters in the string are encoded in 8 bits typically. But we want to see the ones and zeros so we put them in a string

Answer 7

Enter a #: 5
a b c d e

Answer 8

import java.util.Scanner;

//your code goes here
public class Converter{
public static String toBinary(int num)
{
String binary="";
while(num>0){
binary=(num%2)+binary;
num/=2;
}
return binary;
}
}

public class Program {
public static void main(String[ ] args) {
Scanner sc = new Scanner(System.in);
int x = sc.nextInt();
System.out.print(Converter.toBinary(x));
}
}

Answer 9

Thank you! Now it works!

Answer 10

Tq