unsigned right Shift'>>>' Java中的运算符 [英] unsigned right Shift '>>>' Operator in Java

问题描述

可能重复：

为什么（-1>>> 32）= -1？

无符号右移运算符在最左边插入一个0。所以当我这样做时

The unsigned right shift operator inserts a 0 in the leftmost. So when I do

System.out.println(Integer.toBinaryString(-1>>>30))

输出

output

11

因此，它在最左边的位插入0。

Hence, it is inserting 0 in the left most bit.

System.out.println(Integer.toBinaryString(-1>>>32))

输出

output

11111111111111111111111111111111

不应该是0吗？

推荐答案

请参阅 http：// docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.19

如果是推广类型的左侧操作数是int，只有右侧操作数的五个最低位用作移位距离。就好像右手操作数受到按位逻辑AND运算符& （§15.22.1），掩码值为0x1f（0b11111）。因此，实际使用的移位距离始终在0到31的范围内，包括0和31。

If the promoted type of the left-hand operand is int, only the five lowest-order bits of the right-hand operand are used as the shift distance. It is as if the right-hand operand were subjected to a bitwise logical AND operator & (§15.22.1) with the mask value 0x1f (0b11111). The shift distance actually used is therefore always in the range 0 to 31, inclusive.

-1>>> 32 相当于 -1>>> 0 和 -1>>> 33 相当于 -1>>> 1
，尤其令人困惑， -1>>> -1 相当于 -1>>> 31

that is -1 >>> 32 is equivalent to -1 >>> 0 and -1 >>> 33 is equivalent to -1 >>> 1 and, especially confusing, -1 >>> -1 is equivalent to -1 >>> 31

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