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

is this a bug because i am doing it exactly right and it wont recognise my name as a string variable.

Challenge Task 1 of 3

Define a string variable named firstName that stores your name. Set the value to your name.

// I have setup a java.io.Console object for you named console
String FirstName = "Alex" ;

3 Answers

anil rahman
anil rahman
7,786 Points

This is not a bug, it's your variable naming standards:

meaning you have not complied with the camelCase standard of naming your variable.

String firstName = "Alex" ;

thanks for your help anil but why is the "f" in "firstName" got to be in lower case and how will i know when to use captial letters?

anil rahman
anil rahman
7,786 Points

It is the variable naming convention. This is called camelCase.

All variables you create should start with a lowercase character and any word after that should have the first letter capitalised.


String firstName //first letter is lower case and next word has first letter capitalised
String myNameIsAlex //Every other word is capitalised at the first character