Java, comparing BigInteger values

Question

BigInteger bigInteger = ...;


if(bigInteger.longValue() > 0) {  //original code
    //bigger than 0
}

//should I change to this?
if(bigInteger.compareTo(BigInteger.         


        

Answer 1

This is not a direct answer, but an important note about using compareTo().

When checking the value of compareTo(), always test for x < 0, x > 0 and x == 0.
Do not test for x == 1

From the Comparable.compareTo() javadocs:

Compares this object with the specified object for order. Returns a negative integer, zero, or a positive integer as this object is less than, equal to, or greater than the specified object.

Note:

• A negative integer, not -1.
• A positive integer, not 1.

---

True, checking for ==1 and ==-1 would work for BigInteger. This is the BigInteger.compareTo() code:

public int compareTo(BigInteger val) {
    if (signum == val.signum) {
        switch (signum) {
        case 1:
            return compareMagnitude(val);
        case -1:
            return val.compareMagnitude(this);
        default:
            return 0;
        }
    }
    return signum > val.signum ? 1 : -1;
}

But it's still bad practice, and explicitly recommended against in the JavaDocs:

Compares this BigInteger with the specified BigInteger. This method is provided in preference to individual methods for each of the six boolean comparison operators (<, ==, >, >=, !=, <=). The suggested idiom for performing these comparisons is: (x.compareTo(y) <op> 0), where <op> is one of the six comparison operators.

Answer 2

The first approach is wrong if you want to test if the BigInteger has a postive value: longValue just returns the low-order 64 bit which may revert the sign... So the test could fail for a positive BigInteger.

The second approach is better (see Bozhos answer for an optimization).

Another alternative: BigInteger#signum returns 1 if the value is positive:

if (bigInteger.signum() == 1) {
 // bigger than 0
}

Answer 3

If you are using BigInteger, this assumes you need bigger numbers than long can handle. So don't use longValue(). Use compareTo. With your example it better be:

if (bigInteger.compareTo(BigInteger.ZERO) > 0) {

}