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Java Java Basics Getting Started with Java Strings, Variables, and Formatting

call the printf method on the console object and make it write out "<YOUR NAME> can code in Java!" - I don't get this...

Can someone explain what I am doing wrong with this:

compile.printf String firstNameLastName = "Krista Rotchy";

2 Answers

import java.io.Console; public class Introductios { public static void main (String args[]) { Console console=System.console(); String S1=console.readLine("enter your name \n"); console.printf("<%s> can code in java !",S1); System.exit(0); } }

Grigorij Schleifer
Grigorij Schleifer
10,365 Points

Hi Krista,

// I have setup a java.io.Console object for you named console

String firstName; //here you define a new String called firstName
firstName = "Your Name";
  // here you initialise a String value ("Your Name") to firstName
 //initialisation is like storing a value (String) inside a variable
 //a variable represents the value you stored inside youre code
  //firstName is a String variable of your Name now
  //you can also write String firstName = "YourName"; - its the same

console.printf("%s can code in Java ", firstName);

  //the console object is already there, so you dont need to
  //do it like this - Console console = new Console(System.in);
 //In other words, you dont need to create an object of the Console class
 //you will learn about this later on
  //the Console object is called "console" so you use it like this  - console.
  //after calling console use the character(.) to access methods of the Console class
  //one method of the console is printf - so now you have - console.printf
  //here you open the parenthesis and whrite your text inside " "
  //%s is a String representation, where you put your %s, there will be a String like firstName
  //so it looks like this ("bla bla %s ", firstName); , dont forget the (,) and name of your String variable after the " "

Happy coding and stay tuned

Grigorij