instantiation

Possible Duplicate: Creating an “object” of an interface I am new to Java. Based on my understanding: We cannot instantiate an Interface . We can only instantiate a class which implements an interface . The new keyword is used to create an object from a class. However, when I read the source codes of some Java programs, I found that sometimes an Interface is instantiated. For example: Example 1: JButtonObject.addActionListener(new ActionListener() { public void actionPerformed(ActionEvent e) { //codes } }); Example 2: SwingUtilities.invokeLater(new Runnable() { public void run() { //codes } })

Got to know a new thing today that we can create integers by using new operator as below int num = new int(); Now I wonder if I create an integer in this manner then the resulting integer will be a value type or reference type? I guess it will be a value type. I tried the below code int num1 = 10; int num2 = new int(); int num3; num1 = num2; num2 = num3; I got the below build error: Use of unassigned local variable 'num3' I know why this build error is given. But I wonder when and how to use new int() and how exactly does this work? Can anyone please put some light on this? Thanks & Regards :)

Can someone point out to me what I'm doing wrong or where my understanding is wrong? To me, it seems like the code below which instantiates two objects should have separate data for each instantiation. class Node: def __init__(self, data = []): self.data = data def main(): a = Node() a.data.append('a-data') #only append data to the a instance b = Node() #shouldn't this be empty? #a data is as expected print('number of items in a:', len(a.data)) for item in a.data: print(item) #b data includes the data from a print('number of items in b:', len(b.data)) for item in b.data: print(item) if __name_

I've just been experimenting and found that when I run the rolling code, it does not compile and I can't figure out why. My IDE says 'Cannot make a static reference to the non-static field list', but I don't really understand what or why this is. Also what else does it apply to, i.e.: is it just private variables and or methods too and why?: public class MyList { private List list; public static void main (String[] args) { list = new LinkedList(); list.add("One"); list.add("Two"); System.out.println(list); } } However, when I change it to the following, it DOES work: public class MyList {

Is it possible to dynamically instantiate a class using a variable? For example is something like this possible in PHP? class foo { public $something; } $class_name = "foo"; $f = new $class_name(); That should work, yes. You can also do: $f = new $class($arg1,$arg2); timdev Yes, this code will work fine. In PHP 5 can I instantiate a class dynamically? Yes you can, your code should work fine. Yes of course you can instantiate using dynamic names; 来源： https://stackoverflow.com/questions/3112727/in-php-5-can-i-instantiate-a-class-dynamically