int Addition/Subtraction Overflow Detection

[Lambert Stein]

How would you do this EFFICIENTLY?
By that I mean that, if it is done in a tight loop, it would execute fast. Simply checking against the same result done in long doesn't help - it's got the overhead of casting.
Thank!

[Jim Yingst]

[Lambert Stein]: Simply checking against the same result done in long doesn't help - it's got the overhead of casting.

I think this attitude is impractical - anything you do here will be slower than a single unchecked addition or subtraction. You can't reject a possibility because you think it's slower than you'd like - you can only choose the best option you can come up with, whatever its speed. Don't reject an option unless you have a better one in mind.

Having said that, you may get faster results like this:

public static int safeAdd(int x, int y) {
    int z = x + y;
    if (x != z - y)
        throw new ArithmeticException("Overflow adding " + x + " + " + y);
    return z;
}
 
public static int safeSubtract(int x, int y) {
    int z = x - y;
    if (x != z + y)
        throw new ArithmeticException("Overflow subtracting " + x + " - " + y);
    return z;
}

I have not tested this at all, however. I will be interested to hear if you find something faster.
[ October 08, 2007: Message edited by: Jim Yingst ]

[Rob Spoor]

Of course that does not work, since the check will cause an overflow too, therefore the check will pass.

System.out.println(safeAdd(Integer.MAX_VALUE, Integer.MAX_VALUE));
System.out.println(safeSubtract(Integer.MIN_VALUE, Integer.MAX_VALUE));

Output:
-2
1

And you're not going to tell me that there was no overflow here

[Jim Yingst]

Ah, yes. I actually wrote a SafeMath class some years ago and worked through all these, testing carefully. But obviously I don't remember the results as well as I should. (And can't find the class on my computers; oh well.) That trick I used works well with multiplication and division, but not addition and subtraction. All, right, try this:

public static int safeAdd(final int x, final int y) {
    final int z = x + y;
    if (x > 0) {
        if (y > 0 && z < 0)
            throw new ArithmeticException("Overflow adding " + x + " + " + y);
    } else if (y < 0 && z > 0)
        throw new ArithmeticException("Overflow adding " + x + " + " + y);
    return z;
}
 
public static int safeSubtract(final int x, final int y) {
    final int z = x - y;
    if (x > 0) {
        if (y < 0 && z < 0)
            throw new ArithmeticException("Overflow subtracting " + x + " + " + y);
    } else if (y > 0 && z > 0)
            throw new ArithmeticException("Overflow subtracting " + x + " + " + y);
    return z;
}

The condition can be written more compactly, but I've tried to arrange it for maximum efficiency.
[ October 08, 2007: Message edited by: Jim Yingst ]

[Lambert Stein]

So apparently it can't be done any other way than mine, right?
Sure, I guess the previous post will work, but I was trying to avoid 'if' statements.
Thanks to all, though!