Help!!!!!

I am trying to make a program that takes a binary number that the user inputs, and converts it to an int. then it displays it to the user. this is the code i have so far.(it compiles but it doesnt calculate the answer. can anyone help me??

Code:

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import java.util.Scanner;

public class DanielsonErikA2Q2{

        public static void main (String []args){

                String binaryInput;
                int binaryNumber;
                int lastdigit;
                int runningTotal =0;
                int product;
                int preproduct;
                int calculation;
                int result;
                int number;

                Scanner keyboard = new Scanner(System.in);

                System.out.println("Please enter the binary number you would like to convert to a decimal.");

                binaryNumber= keyboard.nextInt();

                while(binaryNumber!=-1){

                while(binaryNumber>0){

                        for(int i=1; i>=100; i++)
                        {
                                                       
                                                        result= (int)Math.pow(2, i);

                                                        number= binaryNumber%10;

                                                        preproduct=  number * result;

                                                        product = number * preproduct;

                                                        runningTotal = runningTotal + product;
                        }
                                                        System.out.println("The Number as a decimal is " + runningTotal);



                }
                }
        }
}

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Quote:

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it doesnt calculate the answer

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Do what *does* the program do when you run it?

Is this for a class assignment? Because there is a much easier way to do this using the parseInt method. Here is an example that you could apply:

Code:

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Integer.parseInt("1001",2);

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It uses a string for the binary number represented. So you could accomplish your task with about 4 lines of code.

It asks for input, then when you put in an answer, it prints out a neverending looop of "the number as a decimal is 0." it doesnt seem to be doing the calculations.

Quote:

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it prints out a neverending looop of "the number as a decimal is 0."

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OK! Now we have something...

The inner while loop (that's the one that most closely contains the output message) begins:

Code:

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while(binaryNumber>0){

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Now the thing is that binaryNumber never changes its value within that loop. SO the loop does whatever it does forever.

You will have to change the logic within that loop.

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Following on from joshdgreen's point: I'm assuming this is an assignment. But do you have to read the user input as an int? Because, really it is not an int: it's a string. Numeric values (like int) don't have a base: numerals (String) do. Even if you don't use the library parse methods you may still find things easier if you deal with the user input as a String and deal with it character by character.

So with the parse int method, could the program calculate any random length number that the user inputs?

Read the API docs for Integer and the BigInteger classes.

Quote:

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Originally Posted by ErikD99

So with the parse int method, could the program calculate any random length number that the user inputs?

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Nope as it has to be a "parsable" integer String, one that represents a valid Integer. If it's greater than Integer.MAX_VALUE or less than Integer.MIN_VALUE, it will throw a NumberFormatException as noted in the Integer API.

Luck!

[edit: oops, as PBrockway links to above]

what would you change the logic of the second while loop then? :s

Quote:

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Originally Posted by ErikD99

what would you change the logic of the second while loop then? :s

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I suppose I would be asking myself what I expect or want the value of binaryNumber to be on the second, third, fourth etc time around the loop. Then after I had used it to calculate runningTotal I would update it so that it had the correct value for the next time around the loop.