Java Long primitive type maximum limit [duplicate]

Answer 1

Exceding the maximum value of a long doesnt throw an exception, instead it cicles back. If you do this:

Long.MAX_VALUE + 1

you will notice that the result is the equivalent to Long.MIN_VALUE.

From here: java number exceeds long.max_value - how to detect?

Answer 2

Long.MAX_VALUE is 9,223,372,036,854,775,807.

If you were executing your function once per nanosecond, it would still take over 292 years to encounter this situation according to this source.

When that happens, it'll just wrap around to Long.MIN_VALUE, or -9,223,372,036,854,775,808 as others have said.

Answer 3

It will overflow and wrap around to Long.MIN_VALUE.

Its not too likely though. Even if you increment 1,000,000 times per second it will take about 300,000 years to overflow.

Answer 4

Ranges from -9,223,372,036,854,775,808 to +9,223,372,036,854,775,807.

It will start from -9,223,372,036,854,775,808

Long.MIN_VALUE.