为什么 Integer.MAX_VALUE + 1 == Integer.MIN_VALUE?

2022-01-14 00:00:00 integer java jls integer-overflow twos-complement

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

是真的.

我知道 Java 中的整数是 32 位的，不能超过 2³¹-1，但我不明白为什么在其 MAX_VALUE 结果中加 1在 MIN_VALUE 而不是某种异常.更不用说像 Ruby 那样将透明转换为更大的类型.

I understand that integer in Java is 32 bit and can't go above 2³¹-1, but I can't understand why adding 1 to its MAX_VALUE results in MIN_VALUE and not in some kind of exception. Not mentioning something like transparent conversion to a bigger type, like Ruby does.

此行为是否在某处指定?我可以依赖它吗?

Is this behavior specified somewhere? Can I rely on it?

推荐答案

因为整数溢出.当它溢出时，下一个值是 Integer.MIN_VALUE.相关JLS

Because the integer overflows. When it overflows, the next value is Integer.MIN_VALUE. Relevant JLS

如果整数加法溢出，则结果是数学和的低位，以某种足够大的二进制补码格式表示.如果发生溢出，则结果的符号与两个操作数值的数学和的符号不同.

If an integer addition overflows, then the result is the low-order bits of the mathematical sum as represented in some sufficiently large two's-complement format. If overflow occurs, then the sign of the result is not the same as the sign of the mathematical sum of the two operand values.

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