简体繁体 English

Java中的逻辑右移运算符

[英]Logical right shift operator in java

原文 2014-10-20 13:50:41 4 2 java

I am beginner to java... I have tried very much but could not find the way the following line 我是Java的初学者...我已经尝试了很多，但是找不到以下行的方式

System.out.println (-1>>>1); System.out.println（-1 >>> 1）;

gives 2147483647 ? 给2147483647？

Can anyone help me ? 谁能帮我？

2 个解决方案

This is because the binary representation of -1 is 11111111111111111111111111111111 . 这是因为-1的二进制表示形式是11111111111111111111111111111111 。 When you perform an unsigned right bit-shift operation ( >>> ) on it it moves all of the bits right by the argument ( 1 in this case) and fills in empty spaces on the left with zeros so you get 01111111111111111111111111111111 which is the binary representation of Integer.MAX_VALUE = 2147483647 (not sure where you got 2147483648 from). 当您在其上执行无符号右移操作（ >>> ）时，它会将所有位向右移至参数（在本例中为1 ），并用零填充左侧的空白区域，从而得到01111111111111111111111111111111 ，即Integer.MAX_VALUE = 2147483647二进制表示形式Integer.MAX_VALUE = 2147483647 （不确定从何处获得2147483648）。

>>> is the bitwise right-shift operator, with 0 sign extension - in other words, all bits "incoming" from the left are filled with 0s. >>>是按位右移运算符，符号扩展为0-换句话说，从左侧“传入”的所有位均填充0。

-1 is represented by 32 bits which are all 1. When you shift that right by 1 bit with 0 sign extension, you end up with a value which has the 31 bottom bits still 1, but 0 for the top bit (the sign bit), so you end up with Integer.MAX_VALUE - which is 2147483647, not 2147483648 as your post states. -1由全部为1的32位表示。当您将其右移1位并加0符号扩展时，最终得到一个值，该值的31个最低位仍为1，但最高位为0（符号位）），因此您最终得到的是Integer.MAX_VALUE -2147483647，而不是帖子所声明的2147483648。

Or in JLS terms, from section 15.19 : 或以JLS的说法，来自第15.19节：

The value of n >>> s is n right-shifted s bit positions with zero-extension, where: n >>> s的值为零扩展的n个右移s位位置，其中：

If n is positive, then the result is the same as that of n >> s . 如果n为正，则结果与n >> s 。

If n is negative and the type of the left-hand operand is int, then the result is equal to that of the expression (n >> s) + (2 << ~s) . 如果n为负，并且左侧操作数的类型为int，则结果等于表达式(n >> s) + (2 << ~s) 。

If n is negative and the type of the left-hand operand is long, then the result is equal to that of the expression (n >> s) + (2L << ~s) . 如果n为负，并且左侧操作数的类型为long，则结果等于表达式(n >> s) + (2L << ~s) 。

This definition ends up being a bit of a pain to work with - it's easier to just work with the "0 sign extension right-shift" explanation, IMO. 使用此定义最终会有些麻烦-仅使用IMO的“ 0符号扩展右移”说明会更容易。