why Integer.MAX_VALUE + 1 == Integer.MIN_VALUE?

Question

System.out.println(Integer.MAX_VALUE + 1 == Integer.MIN_VALUE);

is true.

I understand that integer in Java is 32 bit and can\'t go above 2³

Answer 1

On most processors, the arithmetic instructions have no mode to fault on an overflow. They set a flag that must be checked. That's an extra instruction so probably slower. In order for the language implementations to be as fast as possible, the languages are frequently specified to ignore the error and continue. For Java the behaviour is specified in the JLS. For C, the language does not specify the behaviour, but modern processors will behave as Java.

I believe there are proposals for (awkward) Java SE 8 libraries to throw on overflow, as well as unsigned operations. A behaviour, I believe popular in the DSP world, is clamp the values at the maximums, so Integer.MAX_VALUE + 1 == Integer.MAX_VALUE [not Java].

I'm sure future languages will use arbitrary precision ints, but not for a while yet. Requires more expensive compiler design to run quickly.