Double和Integer的不同溢出策略。 为什么？

Question

this code

System.out.println(Double.MAX_VALUE+12345 == Double.MAX_VALUE);
System.out.println(Integer.MAX_VALUE+12345 == Integer.MAX_VALUE);

returns

true
false

Please clarify this difference.

Answer 1

The rules are the same, it's just that Double.MAX_VALUE is so large that 12345 is roughly 300 orders of magnitude (10 ³⁰⁰ times) smaller. Adding a number that is so much smaller than Double.MAX_VALUE does not change its value. Adding a number that is of the same order of magnitude would make a difference, though:

Double.MAX_VALUE + 1E300

produces a positive infinity result, which is not the same as Double.MAX_VALUE ( demo )

Answer 2

The behavior of addition to the maximum value differs between int and double in two very significant ways:

1. The difference between each int, other than the maximum, and its immediate successor is 1, regardless of magnitude. The difference between consecutive doubles increases with magnitude. For example, the absolute value of the difference between Double.MAX_VALUE and the next double smaller than it is about 2e292. The absolute value of the difference between 1.0 and the next double smaller than it is about 1.1e-16.
2. Every 32-bit bit pattern represents an int number. Some of the 64-bit bit patterns are reserved in double for NaN and infinite values. Adding one to Integer.MAX_VALUE wraps around to Integer.MIN_VALUE . Adding a big enough positive number to make a difference to Double.MAX_VALUE results in Double.POSITIVE_INFINITY , represented by one of the reserved bit patterns.

int is based on 2's complement binary arithmetic. It is designed to be compact, efficient, and to allow very fast simple arithmetic.

double has more of a bias towards functionality. Often, floating point arithmetic takes more than one machine cycle. It can afford to have reserved bit patterns, such as the infinities, that require special handling.