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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.

Name.java
// 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
seal-mask
.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!