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  1. Practice
  2. Java
  3. Introduction
  4. Java Output Formatting
  5. Discussions

Java Output Formatting

  • AmitMaharana + 2 comments
    import java.util.Scanner;

public class Solution {

public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        System.out.println("================================");
        for(int i=0;i<3;i++)
        {
            String s1=sc.next();
            int x=sc.nextInt();
            for(int j=0;j<15;j++){
            if(s1.length()<15){
                s1=s1+" ";
            }
            }
                System.out.print(s1);
            System.out.println(String.format("%03d", x));
        }
        System.out.println("================================");

    }
}
    • Plenoge + 0 comments

      Couple comments for this code. First and foremost, the problem was to test your knowledge of printf.

      Otherwise looking just at your solution, instead of starting at j=0, you could start at s1.length() which would save you from the unnecessary loops where "if(s1.length() < 15)" is false. In fact, with the correcty initialized for loop, you could skip the if statement entirely.

      Also, I wouldn't recommend building up your string in this way (printf aside) as the String is immutable and you're simply wasting memory. Java cannot optimize these appends as they occur on separate lines. You want StringBuilder here instead: http://stackoverflow.com/a/7817989/819373

    • surajnarayan + 0 comments

      System.out.println(String.format("%03d", x)); what will happen if i not write String .format and wat is its use here in this context??