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Java Java Arrays Iteration Looping Like These Programmers

Madeline Yao
Madeline Yao
Full Stack JavaScript Techdegree Student 9,611 Points

Where is the bug

Hello everyone, I am trying to work on the challenge and everything looks alright but it turns out to have some problem and I do not know where is the problem? Could anyone help? Thanks.

public class Programmers {

  public void printMenu() {
    String[] programmers = {
            "Yukihiro Matsumoto",
            "David Nolen",
            "Grace Hopper",
            "Linus Torvalds",

    System.out.println("Choose a programmer:");
    // TODO: Print out a menu by looping through the programmers array.
      The menu should be in the form of (each on a line of its own, starting with 1):
      1. Yukihiro Matsumoto
      2. David Nolen
    for(int i=0;i<programmers.size();i++){




3 Answers

5,604 Points

Hi there! you where pretty close with what you have there, but for printing this output we must use a printf rather than println. The final for loop looking like this:

for(int i=0; i < programmers.length; i++){
   System.out.printf("%d. %s%n", i+1, programmers[i]);

To get the programmers length we must use programmers.length rather than .size(). This is because it is a regular array and not an arraylist. It is easy to get confused between the two of them. For a regular array we use .length and to call the value [i]. For an arraylist we would use .size() and get(i)

We use the %d to represent the integer we need to print, then a period and space, then %s for the string, and a %n to put each output on a newline.

The values that will be displayed in the String are i+1 and the string programmers[i].

If you have any questions don't hesitate to ask. Take care and happy coding :)

Daniel Marin
Daniel Marin
8,021 Points

Hi Madeline Yao , michaelcodes .
So Madeline you were really close. It's good to see that the same problem can be solved a different way.

Ok so here's passing println:

    for(int i=0; i < programmers.length; i++){
      System.out.println(i + 1 + ". " + programmers[i] + "\n");
Mark Casavantes
Mark Casavantes
Courses Plus Student 13,401 Points

Good Morning Madeline and Michael,

I think \n needs to be used instead of %n for a new line in the print statement in Michael's code.