You need to check if a given string contains a valid number, and if so, convert it to binary (internal) form.

Use the appropriate wrapper class’s conversion routine and catch the NumberFormat Exception. This code converts a string to a double :

// StringToDouble.java 
public static void main(String argv[]) { 
    String aNumber = argv[0];    // not argv[1] 
    double result; 
    try { 
        result = Double.parseDouble(aNumber);  
    } catch(NumberFormatException exc) { 
        System.out.println("Invalid number " + aNumber); 
        return; 
    } 
    System.out.println("Number is " + result); 
}

Of course, that lets you validate only numbers in the format that the designers of the wrapper classes expected. If you need to accept a different definition of numbers, you could use regular expressions (see Chapter 4) to make the determination.

There may also be times when you want to tell if a given number is an integer number or a floating-point number. One way is to check for the characters ., d, or e in the input; if it is present, convert the number as a double, otherwise, convert it as an int:

// GetNumber.java 
System.out.println("Input is " + s); 
if (s.indexOf('.') >0 || 
    s.indexOf('d') >0 || s.indexOf('e') >0) 
    try { 
        dvalue = Double.parseDouble(s); 
        System.out.println("It's a double: " + dvalue); 
        return; 
    } catch (NumberFormatException e) { 
        System.out.println("Invalid a double: " + s); 
        return; 
    } 
else // did not contain . or d or e, so try as int. 
    try { 
        ivalue = Integer.parseInt(s); 
        System.out.println("It's an int: " + ivalue); 
        return; 
    } catch (NumberFormatException e2) { 
        System.out.println("Not a number:" + s); 
    } 
}

A more involved form of parsing is offered by the DecimalFormat class, discussed in Section 5.8.