A long bigger than Long.MAX_VALUE

Question

How can I get a long number bigger than Long.MAX_VALUE?

I want this method to return true:

boolean isBiggerThanMaxLong(long val) {     return (val > Long.MAX_VALUE); } 

Answer 1

That method can't return true. That's the point of Long.MAX_VALUE. It would be really confusing if its name were... false. Then it should be just called Long.SOME_FAIRLY_LARGE_VALUE and have literally zero reasonable uses. Just use Android's isUserAGoat, or you may roll your own function that always returns false.

Note that a long in memory takes a fixed number of bytes. From Oracle:

long: The long data type is a 64-bit signed two's complement integer. It has a minimum value of -9,223,372,036,854,775,808 and a maximum value of 9,223,372,036,854,775,807 (inclusive). Use this data type when you need a range of values wider than those provided by int.

As you may know from basic computer science or discrete math, there are 2^64 possible values for a long, since it is 64 bits. And as you know from discrete math or number theory or common sense, if there's only finitely many possibilities, one of them has to be the largest. That would be Long.MAX_VALUE. So you are asking something similar to "is there an integer that's >0 and < 1?" Mathematically nonsensical.

If you actually need this for something for real then use BigInteger class.

Answer 2

You can't. If you have a method called isBiggerThanMaxLong(long) it should always return false.

If you were to increment the bits of Long.MAX_VALUE, the next value should be Long.MIN_VALUE. Read up on twos-complement and that should tell you why.