为什么Integer.MAX_VALUE + 1 == Integer.MIN_VALUE？ [英] why Integer.MAX_VALUE + 1 == Integer.MIN_VALUE?

问题描述

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

is是的。

我知道Java中的整数是32位且不能超过2 ^ 31-1，但我无法理解为什么在MAX_VALUE中加1导致MIN_VALUE，而不是某种异常。没有提到类似透明转换为更大类型的东西，比如Ruby。

I understand that integer in Java is 32 bit and can't go above 2^31-1, 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.

这个行为是否在某处指定了？我可以依赖吗？

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，如某些足够大的二进制补码格式所示。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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