0x80000000在Java中如何等于-2147483648? [英] How is 0x80000000 equated to -2147483648 in java?

问题描述

采用0x80000000的二进制代码，我们得到

Taking the binary of 0x80000000 we get

1000 0000 0000 0000 0000 0000 0000 0000

这等同于-2147483648.我在这个程序中遇到了这个问题.

How does this equate to -2147483648. I got this question with this program.

class a
{
        public static void main(String[] args)
        {
                int a = 0x80000000;
                System.out.printf("%x %d\n",a,a);
        }
}

meow@VikkyHacks:~/Arena/java$ java a
80000000 -2147483648

编辑我了解到2的补码用于表示负数.当我尝试将其与1的补码等同时

EDIT I learned that 2's complement is used to represent negative numbers. When I try to equate this with that 1's complement would be

1's Comp. :: 0111 1111 1111 1111 1111 1111 1111 1111
2's Comp. :: 1000 0000 0000 0000 0000 0000 0000 0000

这又没有任何意义，0x80000000等同于-2147483648

which again does not make any sense, How does 0x80000000 equate to -2147483648

推荐答案

有符号整数溢出会发生这种情况，基本上.

以byte为例更简单. byte值始终在-128到127(含)范围内.因此，如果将值加1，则值为127(即0x7f)，则得到-128.这也是将128(0x80)投射到byte时得到的结果:

It's simpler to take byte as an example. A byte value is always in the range -128 to 127 (inclusive). So if you have a value of 127 (which is 0x7f) if you add 1, you get -128. That's also what you get if you cast 128 (0x80) to byte:

int x = 0x80; // 128
byte y = (byte) x; // -128

溢出(以2s补码整数表示)总是从最高可表达数字到最低可表达数字.

Overflow (in 2s complement integer representations) always goes from the highest expressible number to the lowest one.

对于 unsigned 类型，最大值溢出到0(这也是最低的可表示数字).这在Java中很难显示，因为唯一的无符号类型是char:

For unsigned types, the highest value overflows to 0 (which is again the lowest expressible number). This is harder to show in Java as the only unsigned type is char:

char x = (char) 0xffff;
x++;
System.out.println((int) x); // 0

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