Reputation: 52249
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)
Upvotes: 10
Views: 358
Reputation: 36250
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.
Upvotes: 0
Reputation: 26876
The other reason is that you can do this check yourself very easily and quickly.
if (sum+i < sum) {
throw new AritchmeticException();
}
should do the trick nicely, given that you know i is always positive and less than Integer.MAX_VALUE.
Upvotes: 1
Reputation: 206896
Besides what Jim says, checking for conditions such as overflow would add a performance penalty to any calculation done with integers, which would make programs that do a lot of calculations a lot slower.
Upvotes: 3
Reputation: 3652
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)
Upvotes: 13