java - 如何在一个字节中展开位？

Question

在 Java 中，我有 2 个字节:

byte b1 = (byte) 0b11111111, b2 = (byte) 0b00000000;

我想混合它们，以便每个第一位都来自 b1，而另一个来自 b2(从左到右阅读)。输入的前半部分和后半部分是分开完成的，因此结果是 2 个字节。结果 b3 和 b4 如下所示。

byte b3 = (byte) 0b10101010, b4 = 0b10101010;

为了说明这些位如何是唯一的(使用字母来指定唯一位):

byte b1 = (byte) 0bHGFEDCBA, b2 = (byte) 0bPONMLKJI;

结果将是:

byte b3 = (byte) 0bHPGOFNEM, b4 = 0bDLCKBJAI;

或者，以图形方式，

    +---+---+---+---+---+---+---+---+
b1  | H | G | F | E | D | C | B | A |
    +---+---+---+---+---+---+---+---+
      |   |   |   |   |   |   |   | 
      |   |   |   |   |   |   |   +--------------------------------------------+
      |   |   |   |   |   |   +----------------------------------------+       |
      |   |   |   |   |   +------------------------------------+       |       |
      |   |   |   |   +--------------------------------+       |       |       |
      |   |   |   +-------------------+                |       |       |       |  
      |   |   +---------------+       |                |       |       |       |  
      |   +-----------+       |       |                |       |       |       | 
      +-------+       |       |       |                |       |       |       | 
              |       |       |       |                |       |       |       |  
            +---+---+---+---+---+---+---+---+        +---+---+---+---+---+---+---+---+
         b3 | H | P | G | O | F | N | E | M |     b4 | D | L | C | K | B | J | A | I |
            +---+---+---+---+---+---+---+---+        +---+---+---+---+---+---+---+---+
                  |       |       |       |                |       |       |       |
      +-----------+       |       |       |                |       |       |       |
      |   +---------------+       |       |                |       |       |       |
      |   |   +-------------------+       |                |       |       |       |
      |   |   |   +-----------------------+                |       |       |       |
      |   |   |   |   +------------------------------------+       |       |       |
      |   |   |   |   |   +----------------------------------------+       |       |
      |   |   |   |   |   |   +--------------------------------------------+       |
      |   |   |   |   |   |   |   +------------------------------------------------+
      |   |   |   |   |   |   |   | 
    +---+---+---+---+---+---+---+---+
b2  | P | O | N | M | L | K | J | I |
    +---+---+---+---+---+---+---+---+

实现这一目标的最简单方法是什么？

Answer 1

如果，正如您所说，您一心只想着一句俏话，那么:

public static int interleave(short b1, short b2) {
    return((int)(((b2 * 0x0101010101010101L & 0x8040201008040201L) * 
        0x0102040810204081L >> 49) & 0x5555) |
        (int)(((b1 * 0x0101010101010101L & 0x8040201008040201L) * 
        0x0102040810204081L >> 48) & 0xAAAA));
}

这将返回一个 int，其中 b3 和 b4 作为低 16 位，您可以将其移位并屏蔽掉:

int b3b4 = interleave(b1, b2);
int b3 = b3b4 >> 8;
int b4 = b3b4 & 0b11111111;

算法由Interleave bits with 64-bit multiply提供