How to rearrange bits?

Question

I have to rearagne bits in a byte. I solved the problme like this:

uint8_t c;
uint8_t string[3];

string1[2] = (((c&(1<<0))!=0)<<6)|
             (((c&(1<<1))!=0)<<1)|
             (((c&(1<<2))!=0)<<0)|
             (((c&(1<<3))!=0)<<2)|
             (((c&(1<<4))!=0)<<3)|
             (((c&(1<<5))!=0)<<4)|
             (((c&(1<<6))!=0)<<5)|
             (((c&(1<<7))!=0)<<7);

basicly:

if bit0 is a 1, shift a 1 6times to the left.

if bit1 is a 1, shift a 1 0times to the left. ....

Is there a better solution?

https://stackoverflow.com/questions/10493411/what-is-bit-masking — Alex Reynolds, Sep 20 '17 at 18:32
I would lay out the code in a more readable manner. In its current state it might as well be a PPCG answer. — PeskyToaster, Sep 20 '17 at 18:33
`c` is an *uninitialised variable* so the operation will be undefined behaviour. Please post the typical [Minimal, Complete, and Verifiable example](http://stackoverflow.com/help/mcve) that shows the code. — Weather Vane, Sep 20 '17 at 18:46
Possible duplicate of [reverse bits in byte ansi C](https://stackoverflow.com/questions/22338208/reverse-bits-in-byte-ansi-c). Check the [Bit Twiddling Hacks](http://graphics.stanford.edu/~seander/bithacks.html#BitReverseObvious) page. — vgru, Sep 20 '17 at 18:59
@Groo I deleted my answer which does that, because that is not the problem spec. — Weather Vane, Sep 20 '17 at 19:00
@Groo Not reversing bits, "(c&(1<<1))!=0)<<1)| (((c&(1<<2))!=0)<<0)" end up on the same bit, don't they? — Yunnosch, Sep 20 '17 at 19:02
Yes, just noticed it after the question was reformatted, thanks, removed the vote. — vgru, Sep 20 '17 at 19:03
@Yunnosch, No, If both bits are set, you end up with `(1<<1) | (1<<0)` — ikegami, Sep 20 '17 at 19:05
@ikegami True, just spotted my misthinking, missed the `!=0` everywhere. — Yunnosch, Sep 20 '17 at 19:06

ikegami · Accepted Answer · 2017-09-20T19:34:30.400

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

can also be written as

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

Right off the bat, that eliminates one operation per bit. (Keep in mind that y-x is constant.)

But that's not it. If you apply this transformation throughout, you'll notice that some bits are shifted by the same amount.

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

We can group those.

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

We reduced 30 operations down to 10.

score 0 · Answer 2 · answered Mar 10 '18 at 21:34

"Better" as in: "Beauty is in the eye of the beholder."?

Simply? another way: (uses "no"!? operations other than assignment?)

The 1988 2nd edition of "The C Programming Language" by Kernighan and Ritchie in the last two sections of Chapter 6 explains the subtleties and idiosyncrasies of this code:

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;

Problems? Read the book and consider:

  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;

How to rearrange bits?

2 Answers2