Welcome to the Treehouse Community

Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.

Start your free trial

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

Carolyn Lee
Carolyn Lee
4,118 Points

I keep getting this error on#3: Don't forget to pass your firstName variable as the second argument.

I'm not sure why I keep getting the error. I've got the string identifier in my code, but I don't know what it means to pass it as the 2nd argument.

// I have setup a java.io.Console object for you named console
String firstName = "Carolyn";
console.printf("%s can code in Java!, firstName");

2 Answers

Rob Bridges
.a{fill-rule:evenodd;}techdegree seal-36
Rob Bridges
Full Stack JavaScript Techdegree Graduate 35,467 Points

Hello, very simple fix on this. The console.printf fuction should be

console.printf("%s other words not part of the the variable", variable)

So your's should look like

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

Java is unfortunately very picky, it looks like the quotation mark was just in the wrong place.

Hope this helps!

Carolyn Lee
Carolyn Lee
4,118 Points

Oh...I knew it was something small. Just wasn't seeing it. Thanks!