Why string inputs as doubles are outputting this answer

Question

I'm very new and I'm just curious to know why this code outputs 201.0 when entering 2E2 as the value of the double.

public static void main(String[] args) {
    double r = 2E2;
    try{
    }
    catch(InputMismatchException e) {
        r=-1.0;
    }
    finally{
        r++;
    }
    System.out.println(r);
}

output: 201.0

Because `2E2` is `200.0`. As there is no exceptions in try block only finally block runs and it increases that to `201.0` — alpert, Jun 15 '19 at 05:48

score 3 · Answer 1 · answered Jun 15 '19 at 05:50

3

By putting in the E you are saying 2 * 10 ^ 2 power. The additional 1 comes from the finally block which is always executed even if there is an error thrown. If you were to put 2E3 it would print out 2001.0!

answered Jun 15 '19 at 05:50

Marcus Cantu

463
3
14

score 0 · Answer 2 · answered Jun 15 '19 at 06:13

2E2 in decimal notation is 200. Your try catch block does not contain any code, so it never reaches the exception block and the finally block increments 2E2 by adding 1 to it. So, the answer is 2E2 + 0X1 = 0x2E3 (201.0), which is the correct. Finally, when you print it, it is displayed in the decimal format as 201.0.

Why string inputs as doubles are outputting this answer

2 Answers2