Pass arguments to class constructor

Hello,
Please who can help me to solve my problem? I dynamic
creating class and I want to pass arguments.
I know how can I do creating Function (apply), but I coulnd
find constructor - apply.
var c:Class = Class(getDefinition(className));
var instance:Object;
if(obj){
var args:Array = deserialize(obj);
if(args.length == 1)
instance = new c(args[0]);
else if(args.length == 2)
instance = new c(args[0], args[1]);
else if(args.length == 3)
instance = new c(args[0], args[1], args[2]);
else if(args.length == 4)
instance = new c(args[0], args[1], args[2], args[3]);
else if(args.length == 5)
instance = new c(args[0], args[1], args[2], args[3],
args[4]);
else if(args.length == 6)
instance = new c(args[0], args[1], args[2], args[3],
args[4], args[5]);
else if(args.length == 7)
instance = new c(args[0], args[1], args[2], args[3],
args[4], args[5], args[6]);
else if(args.length == 8)
instance = new c(args[0], args[1], args[2], args[3],
args[4], args[5], args[6], args[7]);
else
return serialize("Unexpected number of arguments");
}else{
instance = new c();
}

Even more follow up advice. Once you have fixed the problem you will then face a new problem.
Test.Inner i1 = t.new Inner();   // compile error
Test.Inner i2 = t.new Inner(1); // okayThis is due to the default constructor not being provided as you have defined your own constructor.

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    ?     Write a program with four methods. One is the main method. The other three are named startMessage, message, and finishMessage.
    ?     Within the startMessage method, place code to print out the message ?The program has started.?
    ?     Within the message method, place code to print out any message that is passed, as a String, into the method using a String parameter named output.
    ?     Within the finishMessage method, place code to print out the message ?The program is done.?
    ?     When the program begins call the startMessage method.
    ?     Then display the following dialog box:
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    ?     Using a While loop, respond to the user?s input appropriately:
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              System.exit(0);
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    ?     The return type of this new method will be a double.
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              double num2 = 0;
              double addResult = 0;
              double subtractResult = 0;
              double multiplyResult = 0;
              double divideResult = 0;
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              addResult = add(num1, num2);
              subtractResult = subtract(num1, num2);
              multiplyResult = multiply(num1, num2);
              divideResult = divide(num1, num2);
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              JOptionPane.showMessageDialog(null, multiplyResult);
              JOptionPane.showMessageDialog(null, divideResult);
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         }//main
         public static double getNumberInput (double input){
              double output = 0;
              Double.parseDouble(JOptionPane.showInputDialog("Enter a number:"));
              return output;
         public static double add (double term1, double term2){
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         }//add
         public static double subtract (double term1, double term2){
              double result = term1 - term2;
              return result;
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         public static double multiply (double term1, double term2){
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              return result;
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         public static double divide (double term1, double term2){
              double result = term1 / term2;
              return result;
         }//divide
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