How to read "0000 0001" from keyboard?

[Vanchi Nathan]

Hello all,

I am trying to read an input from keyboard, but i wanted to use it a a byte data value. When i give "00000001" as input the program throws a NumberFormatException. I have posted the code that i have tried out. Can anyone help in this?

public class TestConversionToDataType {

public static void main (String[] a) {
byte b=0;
String inputString="";

try {
byte[] input = new byte[100];
int count = 0;

System.out.print("Enter a byte value ->");
count = System.in.read(input);

inputString = convertToStringFromBytes(input,count);
i = Integer.parseInt(inputString);
b = Byte.parseByte(inputString,16);
System.out.println(b);
 
} catch (Exception e) {
e.printStackTrace();
}


}

public static String convertToStringFromBytes(byte[] b, int count) {
String temp = new String(b,0,count);
return temp;
}

}
 

Thanks in advance.

[Stephen Huey]

Change your code to this:

try {
    i = Integer.parseInt(inputString);
} catch (NumberFormatException e) {
    System.out.println("Here's the value that caused the exception: " + inputString);
}

Check and make sure that inputString is really a number, because if not, then parseInt won't be happy! Your subject line had a space in between the numbers...I think that would generate an exception.

[siv sanny]

Use this main code

public static void main (String[] a)
{
byte b=0;
String inputString;
try
{
byte[] input = new byte[100];
int count = 0;
System.out.print("Enter a byte value ->");
count = System.in.read(input);
inputString= convertToStrFrBytes(input,count-1);
b = Byte.parseByte(inputString,2);
}
catch (Exception e)
{e.printStackTrace();}


Now try with these changes. and make sure that the binary value you give as input is between -0111111 and 01111111 that is the range of a binary val. that a binary var can hold.

Thanx,
sankarvas