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 IO

4 Answers

Hi there,

There's 4 parts to this question. First is Declare a variable that is named the camel-cased version of "first name". Store the user's first name into this new variable using console.readLine. For that you need to create a String called firstName and ask for user input, as it says. That line looks like:

String firstName = console.readLine("Enter your first name: ");

Next, the question is Declare another variable, naming this one the camel-cased version of "last name". Use console.readLine to store the user's last name into this new variable. THat's pretty much the same as above:

String firstName = console.readLine("Enter your first name: ");
String lastName = console.readLine("Enter your last name: ");

Next you want to output the first name Using the console's printf method, display a message that says, "First name: ", followed by the first name that the user has entered. That's using string interpolation to dd the value held in firstName where the %s is in the output string. Like this:

String firstName = console.readLine("Enter your first name: ");
String lastName = console.readLine("Enter your last name: ");
console.printf("First name: %s", firstName);

And the last tak is similar to the third Using the console's printf method, display a message that says, "Last name: " followed by the last name that the user has entered.

String firstName = console.readLine("Enter your first name: ");
String lastName = console.readLine("Enter your last name: ");
console.printf("First name: %s", firstName);
console.printf("Last name: %s", lastName);

I hope that helps you.

Steve.

thanks a lot steve. it worked perfectly

:+1:

Learning coding
seal-mask
.a{fill-rule:evenodd;}techdegree
Learning coding
Front End Web Development Techdegree Student 9,937 Points

Ok, I am trying to understand the vocabulary here;

console.printf = method? (or function) First name = variable? firstName = parameter?

console.printf("First name: %s", firstName); console.printf("Last name: %s", lastName);

Hi Rene van den Heuvel

Let's try to break this down.

In the first line console is an instance (object) of the Console class. Then, printf is a method (or function) of the Console class instances. A method and function are interchangeable names for the same thing.

Next we are creating a variable of type String called firstName. It could be called an instance of the String class holding the value you assign to it, but let's not. The variable firstName holds a string like "Steve".

In the last line, you are using the string formatting capability of printf. This allows you to insert the value of a variable into a string that you define. You are passing two things into the printf method/function. First is the string you want to have a value inserted into, "First name: %s", then the variable whose value you want to insert at the %s position, firstName. You can call both those things parameters when used in this way. A value passed into a method is a parameter. So, on one line, firstName is a string variable, next you are using the same string variable as a parameter of the printf method.

Let me know if that doesn't make sense and I'll have a better try!

Steve.

Learning coding
seal-mask
.a{fill-rule:evenodd;}techdegree
Learning coding
Front End Web Development Techdegree Student 9,937 Points

Hi Steve,

After reading your answer a few time's I still don't understand it fully, but it helped me understand more. That's good for now. Thanks for your extensive answer.

RenΓ©

If you let me know what's still not clear, I will try again!

Steve.