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Java Java Objects Meet Objects Creating Classes

around 7:59 what did he mean the character name field was exposed?

Also .characterName is a method for strings only right? And it basically just shows what String it is?

public static void main(String[] args) { // Your amazing code goes here... System.out.println("We are making a new PEZ Dispenser"); System.out.printf("FUN FACT: There are %d PEZ allowed in every dispenser %n", PezDispenser.MAX_PEZ); PezDispenser dispenser = new PezDispenser("Yoda"); System.out.printf("The dispenser is %s %n", dispenser.getCharacterName() );

I hope I formatted this correctly, but basically "dispenser" is a reference and Object to the class PezDispenser and we are accessing the member variable of PezDispenser by using a reference to that class Object. that's what he means by exposing the variable. I hope that helps, but that's my take on it. if you have any more questions let me know

1 Answer

Randolph Wang
Randolph Wang
3,614 Points

I had to Google it as well.


"A Java field is a variable inside a class."

From what I understood:

PezDispenser dispenser = new PezDispenser();

That line declares an instance Object (dispenser) of class PezDispenser(). When Craig uses .characterName, it calls the instantiation of the Java field or variable of the class being input as the string literal placeholder. When you refer back to the class file, "Yoda" is set equal to String characterName.