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Inheritance program

Diana Aldueza

Joined: Oct 29, 2002
Posts: 1
Why does:
System.out.println("Please enter the Furlength");
String FurLength = Keyboard.readString();
System.out.println("Please enter the Size of the dog");
String Size = Keyboard.readString();
pets [petCount] = new Dog(Name, Sex, Age, FurLength, Size);
else if(Type='c'){
System.out.println("Please enter the FurColor of the Cat");
String FurColor = Keyboard.readString();
System.out.println("Please enter the EyeColor of the Cat");
String EyeColor = Keyboard.readString();
pets [petCount] = new Cat(Name, Sex, Age, FurColor, EyeColor);
else if(Type='b'){
System.out.println("Please enter the FeatherColor");
String FeatherColor = Keyboard.readString();
System.out.println("Please enter the BeekColor of the Bird");
String BeekColor = Keyboard.readString();
pets [petCount] = new Bird(Name, Sex, Age, FeatherColor, BeekColor);
System.out.println("You have entered the wrong option!");
not work. The error I get says:
but I tried if(Type=='d')
and it still wont work. Can anyone help me find out whats wrong.
Ron Newman
Ranch Hand

Joined: Jun 06, 2002
Posts: 1056
I don't see a declaration for the variable "Type"; what is it?

Ron Newman - SCJP 1.2 (100%, 7 August 2002)
Larry Jones

Joined: Oct 22, 2002
Posts: 24
Looks like Type is a String. To compare strings you need to use the equals() method.
Garrett Smith
Ranch Hand

Joined: Jun 27, 2002
Posts: 401
if(Type='d') is an assignment, so like you figured, you need a boolean statement - if(Type == 'd') is a boolean statement.
Try changing, recompiling, and testing again.

You said you changed = to == .

You got an error that says (Type='d') when you program uses (Type == 'd') -which is probably not what you want if Type is a String
means that you're running the wrong program or forgot to compile.

comp.lang.javascript FAQ:
Ron Newman
Ranch Hand

Joined: Jun 06, 2002
Posts: 1056
Also note that 'd' is a char, while "d" is a String. These are not at all the same thing!
I agree. Here's the link:
subject: Inheritance program
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