java 从 Int 中获取 n 个最低有效位
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Get n Least Significant Bits from an Int
提问by Wingdom
This seems fairly straightforward, but I cant find an answer. If I have an int X, what is the best way to get N least significant bits from this int, in Java?
这看起来相当简单，但我找不到答案。如果我有一个 int X，那么在 Java 中从这个 int 中获取 N 个最低有效位的最佳方法是什么？
回答by Ted Hopp
This should work for all non-negative N < 3332:
这应该适用于所有非负 N < 3332：
x & ((1 << N) - 1)
It's worth elaborating on how this works for N == 31and . For N == 32N == 31, we get 1 << N == Integer.MIN_VALUE. When you subtract 1 from that, Java silently wraps around to Integer.MAX_VALUE, which is exactly what you need. For N == 32, the 1 bit is shifted completely out, so 1 << N == 0; then (1 << N) - 1 == -1, which is all 32 bits set.
值得详细说明这对N == 31和 的作用。对于N == 32N == 31，我们得到1 << N == Integer.MIN_VALUE。当您从中减去 1 时，Java 会默默地环绕到Integer.MAX_VALUE，这正是您所需要的。对于N == 32，1 位被完全移出，所以1 << N == 0; then (1 << N) - 1 == -1，这是所有 32 位设置。
For N == 32, this unfortunately doesn't work because (thanks, @zstring!) the <<operator only shifts by the right side mod 32. Instead, if you want to avoid testing for that case specially, you could use:
对于N == 32，不幸的是这不起作用，因为（谢谢，@zstring！）<<操作员只在右侧 mod 32 处移动。相反，如果您想避免专门针对这种情况进行测试，您可以使用：
x & ((int)(1L << N) - 1)
By shifting a long, you get the full 32-bit shift, which, after casting back to an int, gets you 0. Subtracting 1 gives you -1 and x & -1is just xfor any intvalue x(and xis the value of the lower 32 bits of x).
通过移动 a long，您可以获得完整的 32 位移位，在转换回 an 之后int，得到 0。减去 1 得到 -1，x & -1并且仅x适用于任何int值x（并且x是 的低 32 位的值x）。
回答by Peter Lawrey
Ted's approach is likely to be faster but here is another approach
Ted 的方法可能更快，但这是另一种方法
x << -N >>> -N
This shift all the bit up and then down to chop off the top bits.
这将所有位向上然后向下移动以切掉最高位。
int i = -1;
System.out.println(Integer.toBinaryString(i));
i = i << -5 >>> -5;
System.out.println(Integer.toBinaryString(i));
prints
印刷
11111111111111111111111111111111
11111
回答by user12042
You can also use a mask. If you use the & bitwise operator you can then remove whatever bit you would want to remove (say the highest x bits);
您也可以使用面膜。如果您使用 & 按位运算符，则可以删除您想要删除的任何位（例如最高的 x 位）；
int mask = 0x7FFFFFFF //Example mask where you will remove the
// most significant bit
// (0x7 = 0111b and 0xF = 1111b).
int result = numberToProcess & mask; //And apply the mask with the &bitwise op.
The disadvantage to this is that you will need to make a mask for each bit, so perhaps this is better seen as another method of approach in general.
这样做的缺点是您需要为每个位制作一个掩码，因此通常将其视为另一种方法更好。
