如何重新排列位?

(((c&(1<<x))!=0)<<y)

也可以写成

((c&(1<<x)))<<(y-x))

一开始,这就消除了每比特一个操作。(请记住 y-x

但事实并非如此。如果在整个过程中应用此转换,您会注意到一些位的移位量相同。

(( c & 0x01 ) << 6 ) |
(( c & 0x02 )      ) |
(( c & 0x04 ) >> 2 ) |
(( c & 0x08 ) >> 1 ) |
(( c & 0x10 ) >> 1 ) |
(( c & 0x20 ) >> 1 ) |
(( c & 0x40 ) >> 1 ) |
(( c & 0x80 )      )

我们可以将它们分组。

(( c & 0x01 ) << 6 ) |
(( c & 0x82 )      ) |
(( c & 0x78 ) >> 1 ) |
(( c & 0x04 ) >> 2 )

“更好”如:“情人眼里出西施。”?

Kernighan和Ritchie在1988年出版的第2版《C编程语言》第6章最后两节中解释了这段代码的微妙之处和特点:

uint8_t c;
uint8_t string[3];

union { uint8_t  bits;
  struct { uint8_t t7    :1;
           uint8_t t3456 :4;
           uint8_t t2    :1;
           uint8_t t1    :1;
           uint8_t t0    :1;
         } bi;
  struct { uint8_t b7    :1;
           uint8_t b6    :1;
           uint8_t b2345 :4;
           uint8_t b1    :1;
           uint8_t b0    :1;
         } ti;
      } b,d;

  b.bits = d.bits = c;      // does b.ti.b1 = d.bi.t1; b.ti.b7 = d.bi.t7;

  b.ti.b0    = d.bi.t2;
  b.ti.b2345 = d.bi.t3456;
  b.ti.b6    = d.bi.t0;

  string[2] = b.bits;

  struct { uint8_t t0    :1;
           uint8_t t1    :1;
           uint8_t t2    :1;
           uint8_t t3456 :4;
           uint8_t t7    :1;
          } bi;
  struct { uint8_t b0    :1;
           uint8_t b1    :1;
           uint8_t b2345 :4;
           uint8_t b6    :1;
           uint8_t b7    :1;
          } ti;