Reputation: 126
If I have an assignment
Long c = a + b;
Is there an easy way to check that a + b
is not bigger/smaller than Long.MAX_VALUE
/Long.MIN_VALUE
?
Upvotes: 9
Views: 2526
Reputation: 198113
Using Guava, it's as simple as
long c = LongMath.checkedAdd(a, b); // throws an ArithmeticException on overflow
which is, I'd like to think, very readable indeed. (LongMath Javadoc here.)
For the sake of fairness, I'll mention that Apache Commons provides ArithmeticUtils.addAndCheck(long, long)
.
If you want to know how they work, well, the answer is one line of bit-hackery for Guava: the result doesn't overflow if (a ^ b) < 0 | (a ^ (a + b)) >= 0
. This is based on the trick that the bitwise XOR of two numbers is nonnegative iff they have the same sign.
So (a ^ b) < 0
is true if a
and b
have different signs, and if that's the case it'll never overflow. Or, if (a ^ (a + b)) >= 0
, then a + b
has the same sign as a
, so it didn't overflow and become negative.
(For more tricks like this, investigate the lovely book Hacker's Delight.)
Apache uses more complicated casework based on the sign of a
and b
.
Upvotes: 17
Reputation: 48196
the simple route:
if(a/2+b/2+(a&b&1)>long.MAX_VALUE/2||a/2+b/2<long.MIN_VALUE/2)...
you just need to hope that it doesn't get optimized to (a+b)/2
Upvotes: 0
Reputation: 1074455
It's only an issue if they have the same sign (and are both !0
), since otherwise you're safe from overflow. If overflow occurs, the sign of the result will flip. So:
long r = a + b;
if ( (a < 0 && b < 0 && r >= 0) ||
(a > 0 && b > 0 && r <= 0) ) {
// Overflow occurred
}
Upvotes: 13
Reputation: 372814
One option would be to use the BigInteger
class to do the exact computation, then check whether the result is greater than or smaller than the value in question. For example:
if (BigInteger.valueOf(a).add(BigInteger.valueOf(b)).compareTo(BigInteger.valueOf(Long.MAX_VALUE) > 1) {
/* Overflow occurred. */
} else {
/* No overflow occurred.
}
Hope this helps!
Upvotes: 0