Thread: Java has a default constructer, so why there is a mistake at

Hi

It is in Java.

class fd {
int x;
void generator (int i){ fd o=new fd (i);}
}

class ibj2 {
public static void main (String args[]){
fd ob=new fd ();
for (int j=0; j <=10; j++) ob.generator (j);
}
}

It is supposed that x=0 automatically since Java has a default constructer.

First of all, it would be useful to use proper code tags to preserve syntax highlighting.
Second, if you could show us the exact error you get it will become much easier for us to give you a proper answer. Right now, we have to guess what the error is.

I guess the error comes because your class "fd" does not declare any constructors, but you are trying to use a non-default constructor in your code.
The line in question is:

void generator (int i){ fd o=new fd (i);}

Here you try to pass the argument "i" to the constructor. But the default constructor does not take any parameters.

when I compiled the program, this is what I got as error:-

constructor fd in class fd cannot be applied to the given types;
required: no arguments;
found: int;

If you still didn't understand, then let me show you:-

fd o=new fd (i);

Here you cannot write i in the brackets () as "(i)". You can write

fd o=new fd ();

but that "might" defeat the purpose of your making the program. I'm not sure if it actually will.

Thank you for asking.

Thanks a lot to all of you

--- Update ---

I just thought that the j variable of for, would has been put directly in the generator method, it seems that it must be put into the instance variable x first. Am I wrong?

--- Update ---

I am just trying to understand the mechanism of it.

Thanks.

I have read that the contents of a String object are immutable, but I could change them as I wish, check this out:

class a {
public static void main (String args []) {
String x= "mummy"; System.out.println (x);
x="daddy"; System.out.print (x);

}
}

As I see, there is no point of knowing that they cannot be altered since I can change the contents.
Could anyone tell me what I have missed here?

--- Update ---

Is there any new updates from Oracle that has applied on String class?

--- Update ---

Plus, how can I retrieve a reference variable for a string object? As I do for an variable of a class.

--- Update ---

* a variable of a class type

I think you are confusing objects and variables.
In your example you have one variable, it is called x.
But you have 2 different String objects. One of these Strings is "mummy" and the other String is "daddy".
You assign these objects to your variable, but the Strings themselfs never change. Only the reference of the variable x changes.

After

x="daddy";

the String "mummy" still exists, and it still reads mummy.

If you want to make sure that the reference of your variable x will never change you can declare it final. For example:

final String x = "mummy";

Now you could not assign a different value to this variable anymore.

Works for me:

public class Test {
 
	public static void main(String[] args) {
		String s = new String();
		Integer i = new Integer(42);
 
		System.out.println(s+"\n"+"i");
	}
}

As I said, you can print it.
The program I posted above works like a charm.

If you are getting an error, please provide it to us.

Cornix: I have two classes in my code.

--- Update ---

My code has no error. Please, understand my point. y and z are both references to objects but I can only print one of them.

So, y is not a reference variable? It doesn't have an address for an object? Seems like the normal variables. If you excute my code you will see my point.

Originally Posted by magDreameR

So, y is not a reference variable? It doesn't have an address for an object? Seems like the normal variables. If you excute my code you will see my point.

y is a reference to a String object in your code. And I am sure that you can print it.
But the y in your example code is an empty String. So perhaps you think that you cant print it because you dont see anything.