Long + Long not bigger than Long.MAX_VALUE

问题

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?

回答1:

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.

回答2:

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
}

回答3:

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!

回答4:

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

来源：https://stackoverflow.com/questions/9057290/long-long-not-bigger-than-long-max-value

标签

java

long-integer