Throw Exception in this code

[eng_hyzoom]

in the following code converts Binary String into Decimal number
i want to throw NumberFormatException if the string isnt binary
how can i do that ?

public class Exercise8_8 {
  public static void main(String[] args) {
    // Prompt the user to enter a string
    java.util.Scanner input = new java.util.Scanner(System.in);
    System.out.print("Enter a binary number string: ");
    String s = input.nextLine();
    System.out.println("The decimal value is " + parseBinary(s));
  }

  public static int parseBinary(String binaryString) {
    int value = binaryString.charAt(0) - '0';
    for (int i = 1; i < binaryString.length(); i++) {
      value = value * 2 + binaryString.charAt(i) - '0';
    }

    return value;
  }
}

Thanks...

[SmilingKey]

You can use Try Catch

[eng_hyzoom]

yes i know
how can i do that?
thanks for reply.

[Fubarable]

You can find out how to do this from the tutorials. Please have a look here: Exceptions Tutorial

[eng_hyzoom]

thanks all for reply
i under stand the concept but i steal cant use it
and this code is home work
so i want to do it before tomorrow
thanks again

[JosAH]

thanks all for reply
i under stand the concept but i steal cant use it
and this code is home work
so i want to do it before tomorrow
thanks again

If you detect a situation where you want to throw that exception, simply throw it:

throw new NumberFormatException("you goofed!");

... and let anything else catch it and do with it what it wants.

kind regards,

Jos

[eng_hyzoom]

thanks very much Jos
this method parseBinary

public static int parseBinary(String binaryString) {
    int value = binaryString.charAt(0) - '0';
    for (int i = 1; i < binaryString.length(); i++) {
      value = value * 2 + binaryString.charAt(i) - '0';
    }

it used to convert binary String into binary
i want to put exception if user enter String not binary String
i dont know where i put this exception i tried put it was error
this is my trial there are 5 errors

 public static int parseBinary(String binaryString) {
	  try{
		  int value = binaryString.charAt(0) - '0';
		      for (int i = 1; i < binaryString.length(); i++) {
      value = value * 2 + binaryString.charAt(i) - '0';
	  }
   	  catch(Exception ex){
		  System.out.println("Exception");
	  }
    }
    return value;
  }

[Eranga]

Can you please use code tags nest time.

[Eranga]

What you mean by there are 5 errors?

What you have to do is, if you comes with an error return a value that you can handle in the invoked method, main in your case.

[JosAH]

thanks very much Jos
this method parseBinary

public static int parseBinary(String binaryString) {
    int value = binaryString.charAt(0) - '0';
    for (int i = 1; i < binaryString.length(); i++) {
      value = value * 2 + binaryString.charAt(i) - '0';
    }

it used to convert binary String into binary
i want to put exception if user enter String not binary String

Your value 'binaryString.charAt(i) - '0'' needs to be a value 0 or 1; if it isn't simply throw that exception just as I showed you in my previous reply:

int v= binaryString.charAt(i) - '0';
if (v < 0 || v > 1)
   throw new NumberFormatException("you goofed!");

kind regards,

Jos