Java整数Min_Value为负，然后比较 [英] Java Integers Min_Value negative then compare

问题描述

明天我有一个测试，我无法理解我的书籍解释，我很感激帮助：

I have a test tomorrow and I cant understand my books explanation, I appreciate the help:

public class TestClass{
      public static void main(String[] args) throws Exception{
            int a = Integer.MIN_VALUE;
            int b = -a;
            System.out.println( a+ "   "+b);
      }
}

输出： -2147483648 - 2147483648

为什么打印2个相同幅度的负数而不是正数和负数？

Why does this print 2 negative numbers of the same magnitude and not a positive and negative?

推荐答案

由于无提示整数溢出： Integer.MIN_VALUE 是 -2 ^ 31 和 Integer.MAX_VALUE 是 2 ^ 31-1 ，所以 -Integer.MIN_VALUE 是 2 ^ 31 ，这是 Integer.MAX_VALUE + 1 ，根据定义，它对于整数来说太大了。所以它溢出并变成 Integer.MIN_VALUE ...

Because of silent integer overflow: Integer.MIN_VALUE is -2^31 and Integer.MAX_VALUE is 2^31-1, so -Integer.MIN_VALUE is 2^31, which is Integer.MAX_VALUE + 1, which by definition is too large for an integer. So it overflows and becomes Integer.MIN_VALUE...

您还可以检查：

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

打印相同的东西。

更多从技术上讲，结果由 Java语言定义规范＃15.18.2 ：

More technically, the result is defined by the Java Language Specification #15.18.2:

如果整数加法溢出，则结果是数学的低位以一些足够大的二进制补码格式表示的和。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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