某些按位操作可以解决问题.

Some bitwise operations can do the trick.

#include <inttypes.h>

int main(){

  uint8_t rows[8] = {
      0b11111111,
      0b00000001,
      0b00000001,
      0b00111111,
      0b00000001,
      0b00000001,
      0b00000001,
      0b11111111
  };


  uint8_t rows2[8] = {0, 0, 0, 0, 0, 0, 0, 0};
  uint8_t rows3[8] = {0, 0, 0, 0, 0, 0, 0, 0};

  int i, j;
  // rotate clockwise
  for(i=0; i<8; ++i){
    for(j=0; j<8; ++j){
      rows3[i] = (  ( (rows[j] & (1 << (7-i) ) ) >> (7-i) ) << j ) | rows3[i];
    }
  }

  // rotate anti-clockwise
  for(i=0; i<8; ++i){
    for(j=0; j<8; ++j){
      rows2[i] = (  ( (rows[j] & (1 << i ) ) >> i ) << (7-j) ) | rows2[i];
    }
  }
}

在顺时针方向上，使用(rows[j] & (1 << (7-i) ) ) >> (7-i)获取第j个原始字节的第(7-i)个位，然后将其移至第j个位置.通过对字节本身执行或"(|)来收集所有位，因此将数组初始化为0非常重要. 逆时针情况类似，更改了索引. 我用另一封信对其进行了测试，可以让您确定旋转是否正常工作.如果您需要进一步的解释，请问.

In the clockwise case, you get each (7-i)-th bit of the j-th original byte with (rows[j] & (1 << (7-i) ) ) >> (7-i) and then shift it to the j-th position. You collect all the bits by doing an "or" (|) with the byte itself, so it is very important to initialize the array with 0s. The anti-clockwise case is analogous, changing the indexing. I used another letter to test it, that let you know for sure if the rotation is working properly. If you need further explanation, please just ask.

如果要查看结果，则在此SO问题中使用该函数:

If you want to look the result, I'm using the function in this SO question: Is there a printf converter to print in binary format?