增加primitve类型的MAX_VALUE

Question

I try to understand the following code:

int i = Integer.MAX_VALUE;
double d = Double.MAX_VALUE;

System.out.println(i + ":" + (i+1));
System.out.println(d + ":" + (d+1));

The output is:

2147483647:-2147483648
1.7976931348623157E308:1.7976931348623157E308

For the first line, i is equal to the maximum integer and incrementing it leads to the lowest integer value. Why the same is not happening with d ?

Is there a simple explanation for this behaviour?

Answer 1

When a double overflows, its value becomes Infinity , not a negative number.

The reason why you can't overflow by simply adding 1 to Double.MAX_VALUE is that MAX_VALUE + 1 == MAX_VALUE because the precision of doubles is not enough to make the difference between those two numbers.

Actually you can add quite a large number before jumping to the next available double. For example that does not overflow either:

System.out.println(Double.MAX_VALUE + 1e100); //still Double.MAX_VALUE

This overflows as expected:

System.out.println(m + (Math.ulp(m) / 2)); //Infinity

Answer 2

Your double cannot represent the actual number you're trying to represent due to lose of accuracy with double s. That's why the value is not changing and you're actually don't have an overflow.

For non- int types, overflow will result in Infinity . Try to multiply that value with 20, for example, and you'll see the overflow in action.

Answer 3

The integer overflows because it is represented in a two's complement format.

• Integer.MAX_VALUE is represented by 0111...111
• Integer.MIN_VALUE is represented by 1000...000

So when you try and add 1 to Integer.MAX_VALUE , according to the normal rules of addition, you get Integer.MIN_VALUE . This is called overflow.

Floating point numbers have a different representation, which doesn't lend itself to this type of overflow. They are all approximations, and the closest approximation it has to the value you gave it is Double.MAX_VALUE .