Java为什么要砍掉这个长整数的高阶位？

Question

Consider the following code:

System.out.println(1 + 0xFFFFFFFFL);
System.out.println(1L + 0xFFFFFFFF);

The first line prints the expected value, 4294967296 . But the second line prints a 0 . I checked the type of both expressions (by passing them to methods) and both are recognized as long by the JVM. The Lava 7 language specification states that with binary operations, "if either operand is of type long, the other is converted to long." It seems that's what is happening, but I have two questions:

1. If both operands end up being longs, why are the higher-order bits chopped off in the first expression?

2. Why should the order matter?

Answer 1

0xFFFFFFFF is equal to -1 . When you add 1 + -1 you get 0.

The order matters because 0xFFFFFFFFL != (long) 0xFFFFFFFF

Just like (double) 0.1F != 0.1

Answer 2

By default, in java all numeric constants are int .

In the second example, because java uses the two's-complement binary representation, the expression 0xFFFFFFFF is -1 as an int , which when widened to a long stays as -1 , so you've coded 1 + -1 , giving zero.