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

Anurag Gracian
Anurag Gracian
315 Points

i do not know why my string variable is not working

i defined the string variable. but the code is not working

// I have setup a java.io.Console object for you named console
String firstName = "Anurag";
  console.printf("%s\n, firstName");


1 Answer

Hey there Anurag!

Your code is almost perfect.

The console.printf() function requires one extra argument for every formatter you pass into it. Take this as an example:

console.printf("%s %s %s %s");
//This will throw an error.

You have 4 string formatters, which means you must have 4 strings (either as literals or variables) passed into the method.

The code, when properly written, would look like this:

String one = "one";
String two = "two";
String three = "three";
String four = "four";

console.printf("%s %s %s %s", one, two, three, four);
//The console will display: one two three four

For your code specifically, you need to remove "firstName" from the string and rewrite it like so:

// I have setup a java.io.Console object for you named console
String firstName = "Anurag";
  console.printf("%s\n", firstName);


This is telling the printf() function that you have a string ("%s\n") with one variable/literal being passed into it (firstName).

Hopefully this helps. :-)