The answer is in the question. If a floating point number literal ends with F of f, it's a float. Otherwise, it's a double.
It is valid, but you may get interesting results in edge cases if you don't specify a precision on the double... With the caveat from @PinnyM, I should point out that converting int to double is lossless, but promoting long to double is lossy.
Try this:
if (items.elementAt(1) instanceof Double) {
sum.add( i, items.elementAt(1));
}
Since this is the first question from Google I'll add the JavaScript style typeof
alternative here as well:
myObject.getClass().getName() // String
Reflection is slower, but works for a situation when you want to know whether that is of type Dog or a Cat and not an instance of Animal. So you'd do something like:
if(null != items.elementAt(1) && items.elementAt(1).getClass().toString().equals("Cat"))
{
//do whatever with cat.. not any other instance of animal.. eg. hideClaws();
}
Not saying the answer above does not work, except the null checking part is necessary.
Another way to answer that is use generics and you are guaranteed to have Double as any element of items.
List<Double> items = new ArrayList<Double>();
Use regular expression to achieve this task. Please refer the below code.
public static void main(String[] args) {
try {
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
System.out.print("Enter your content: ");
String data = reader.readLine();
boolean b1 = Pattern.matches("^\\d+$", data);
boolean b2 = Pattern.matches("[0-9a-zA-Z([+-]?\\d*\\.+\\d*)]*", data);
boolean b3 = Pattern.matches("^([+-]?\\d*\\.+\\d*)$", data);
if(b1) {
System.out.println("It is integer.");
} else if(b2) {
System.out.println("It is String. ");
} else if(b3) {
System.out.println("It is Float. ");
}
} catch (IOException ex) {
Logger.getLogger(TypeOF.class.getName()).log(Level.SEVERE, null, ex);
}
}
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