The Range of an int and Overflow

An int in Java is always 32 bits, no more, no less. Thirty-two bits can represent exactly 2³² different patterns, and Java splits that range so roughly half the patterns represent negative numbers and half represent nonnegative ones. That gives a ceiling, Integer.MAX_VALUE, which is 2,147,483,647, and a floor, Integer.MIN_VALUE, which is -2,147,483,648. There is no bit pattern left over for 2,147,483,648. The int does not expand to make room, because its size is fixed at compile time.

So what happens when you add 1 to the biggest int? The bits just do normal binary addition, and the carry that would need a 33rd bit falls off the end and is thrown away. The leftover pattern is read back using the same rule that made MAX_VALUE positive and MIN_VALUE negative, and that pattern happens to mean -2,147,483,648. Java does not throw an error or warn you. The program keeps running with a wildly wrong value, which is exactly why overflow bugs are dangerous: nothing looks broken until the numbers do.

This matters right after you meet variables and expressions, because it is the first place a seemingly correct line of code gives a wrong answer for a reason that has nothing to do with logic and everything to do with storage. The fix, when a value might exceed about 2 billion, is to switch to a type with more room, like double or long, trading away the fixed 32-bit ceiling for a much wider range, at the cost of some precision for double.