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Java Java Arrays Iteration Looping Like These Programmers

Michelle Huang
Michelle Huang
5,261 Points

Compiler error with %n. Isn't that supposed to create a new line?

it's saying that it can't find the symbol %n

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.length; i++ )
        System.out.printf((i+1) + ". " + programmers[i] %n);



2 Answers

Gergely Horvath
Gergely Horvath
Courses Plus Student 8,207 Points

I'd like to point out use cases of printf and println: You user println for (concatenated) string, it auto-adds a newline char:

System.out.println((i+1) + ". " + programmers[i]);

printf prints out formatted string, and it accepts multiple arguments:

System.out.printf("%d. %s%n", (i+1), programmers[i]);

In this case in the first argument you specify string format, then you add the corresponding variables in the following arguments.

Michelle Huang
Michelle Huang
5,261 Points

I see, so for printf, you have to use variables for it to work and no concatenation is accepted?

Gergely Horvath
Gergely Horvath
Courses Plus Student 8,207 Points

You don't have to use variables, and it accepts concatenation, but it's nice to extract variables from the representation format.

Jerson Otzoy
Jerson Otzoy
1,532 Points

Hello, This will fix it, you were missing the + sign to concatenate the new line, and the new line is "\n" instead of %n

System.out.printf((i+1) + ". " + programmers[i] + "\n");