Why is Java not telling me when I can’t use Integer?

Question

Hello

For a small project (Problem 10 Project Euler) i tried to sum up all prime numbers below 2 millions. So I used a brute force method and iterated from 0 to 2'000'000 and checked if the number is a prime. If it is I added it to the sum:

private int sum = 0;

private void calculate() {
   for (int i = 0; i < 2000000; i++) {
      if (i.isPrime()) {
         sum = sum + i;
      }
   }
   sysout(sum)
}

The result of this calculation is 1179908154, but this is incorrect. So i changed int to BigInteger and now i get the correct sum 142913828922. Obviously the range of int was overflowed. But why can't Java tell me that? (e.g. by an exception)

Answer 1

Because it's conceivable that you might want it to behave in the traditional Integer fashion. Exceptions are reserved for things that are definitely and irrevocably wrong.

ETA: From the language spec:

    "The built-in integer operators do not indicate overflow or underflow in any way. The only numeric operators that can throw an exception (§11) are the integer divide operator / (§15.17.2) and the integer remainder operator % (§15.17.3), which throw an ArithmeticException if the right-hand operand is zero."

(http://java.sun.com/docs/books/jls/second_edition/html/typesValues.doc.html)

Answer 2

Because our profession values performance over correctness. ;(

Using BigInteger by default, and only reasoning whether it is acceptable, to use long or int if performance is a real problem, would help to avoid such problems.