'Print loop output horizontally?

Hey guys I have to write a program for class that asks me to produce a bar graph based on how many cars a salesperson sold for the month. An example is this:

Pam XXXX

Leo XXXXXX

I'm almost there but not quite. As of right now I can get one X next to the salespersons name but any others are printed underneath on separate lines.

Pam X 

X


X

If you can help me get those other X's on the same line I would really appreciate it. Thank you for your input!

import java.util.Scanner;
public class BarGraph
{
    public static int Pam = 0;
    public static int Leo = 0;
    public static int Kim = 0;
    public static int Bob = 0;

    public static Scanner kb = new Scanner(System.in);

    public static void main(String args[])
    {
        System.out.println("Enter number of cars sold by Pam ");
        Pam = kb.nextInt();
        kb.nextLine();
        System.out.println("Enter number of cars sold by Leo ");
        Leo = kb.nextInt();
        kb.nextLine();
        System.out.println("Enter number of cars sold by Kim ");
        Kim = kb.nextInt();
        kb.nextLine();
        System.out.println("Enter number of cars sold by Bob ");
        Bob = kb.nextInt();
        kb.nextLine();

        System.out.print("Pam "); pamsCars();
        System.out.print("Leo "); leosCars();
        System.out.print("Kim "); kimsCars();
        System.out.print("Bob "); bobsCars();
    }

    public static void leosCars()
    {
        for(int i=0; i < Leo; i++)
        {
            printX();
        }
    }

    public static void kimsCars()
    {
        for(int i=0; i < Kim; i++)
        {
            printX();
        }
    }

    public static void pamsCars()
    {
        for(int i=0; i < Pam; i++)
        {
            printX();
        }
    }

    public static void bobsCars()
    {
        for(int i=0; i < Bob; i++)
        {
            printX();
        }
    }

    public static void printX()
    {
        System.out.println("X");
    }
}
javafor-loop


Solution 1:[1]

You are using System.out.println("X"); This automatically appends a newline character to the end of the string.

Instead use System.out.print("X"); to print the X's next to each other.

Solution 2:[2]

couldn't you use a 

System.out.printf("%s",x);

as well?

or 

System.out.printf("%n%s",x);

i didn't try the loop myself but i'm pretty sure one of the two would also be an alternative... the %n will give you the same result as println as long as you go with printf and %n... its the same escape sequence as \n when you're using println.

i just learned that a month ago in class, so i figured i'd share what basic knowledge i have... mid-terms next week so i have programming fever.

Solution 3:[3]

use print instead of println in your printX function to be specific.

since println prints and adds break (newline)

Solution 4:[4]

public static void printX()
    {
        System.out.println("X");  // println() puts newline character.. use print()
    }

Solution 5:[5]

Change your printX() method not to print newline and change the printCar() signature to accept name and number of X to print.

public static void printX()
{
    System.out.print("X");
}

public static void printCars(String name, int num)
{
    System.out.print(name);
    for(int i=0; i < num; i++)
    {
        printX();
    }
    System.out.println();
}

Now you can call them like 

printCars("Pam ",Pam));
printCars("Bob ",Bob));

Solution 6:[6]

make these changes: 

 System.out.print("Pam "); pamsCars();System.out.println("");

and 

 public static void printX()
    {
        System.out.print("X");
    }

As the others have said, the "println()" method adds a "\n" to the end of the line.

Solution 7:[7]

How about the following (sorry not tested, so in principle only and more of a meansd of shjowing an alternative to loops).

Of course a flaw is that a user could input more than 10.

There again what if the user input 10,000? (would you really want to print 10,000 X's). Perhaps something to think about.

import java.util.Scanner;
public class BarGraph
{
    public static int Pam = 0;
    public static int Leo = 0;
    public static int Kim = 0;
    public static int Bob = 0;

    public static Scanner kb = new Scanner(System.in);

    //Assume max sales of 10
    public static final String maxsales = "XXXXXXXXXX"; //+++++NEW

    public static void main(String args[])
    {
        System.out.println("Enter number of cars sold by Pam ");
        Pam = kb.nextInt();
        kb.nextLine();
        System.out.println("Enter number of cars sold by Leo ");
        Leo = kb.nextInt();
        kb.nextLine();
        System.out.println("Enter number of cars sold by Kim ");
        Kim = kb.nextInt();
        kb.nextLine();
        System.out.println("Enter number of cars sold by Bob ");
        Bob = kb.nextInt();
        kb.nextLine();

        printCars("Pam",pam);  //+++++NEW
        printCars("Leo",leo);  //+++++NEW
        printCars("Kim",kim);  //+++++NEW
        printcars("Bob",bob);  //+++++NEW

    }

    //+++++NEW All other methods/functions replaced by this one
    public static void printCars(String person, int sales) {
        system.out.println(person + " " + maxsales.substr(sales));
    }
}

Solution 8:[8]

// Incrementing by one in Java
for (int i=0; i<10; i++) {
    System.out.println(i);
}
        
// Incrementing by one
// Printing Horizontally
for (int i=0; i<10; i++) {
    System.out.print(" " + i);
}

Solution 9:[9]

int n = 20;
for(int i = 0; i <= n; i++) {
    System.out.print(" " + i);
}

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 Phil Schwartz
Solution 2 sneufeld
Solution 3 Saher Ahwal
Solution 4 TheLostMind
Solution 5 anonymous
Solution 6
Solution 7 MikeT
Solution 8 Tyler2P
Solution 9 Cerbrus

Related Questions