如何循环移位的4字符数组？ [英] How to circular shift an array of 4 chars?

问题描述

我有四个无符号的字符数组。我要像对待一个32位数字（假设焦炭的高位不在乎。我只在乎低8位）。然后，我想通过循环地方任意数量转向它。我有几个不同尺寸的转变，都在编译时确定的。

例如

  unsigned char型A [4] = {0×81，为0x1，为0x1，0X2};
circular_left_shift（一，1）;
/ *一个是现在{0X2，0X2，0X2，0x5的} * /
 

编辑：！？为了大家奇怪，为什么我没有提到CHAR_BIT = 8，因为这是标准C.我没有指定一个平台，那你为什么假设有一个

解决方案

 静态无效rotate_left（uint8_t有* D，uint8_t有* S，uint8_t有位）
{
   常量uint8_t有octetshifts =比特/ 8;
   常量uint8_t有bitshift位= 8％;
   常量uint8_t有bitsleft =（8  -  bitshift）;
   常量uint8_t有流明=（1 <<;＆下; bitshift） -  1;
   常量uint8_t有UM =〜流明;
   INT I;   对于（I = 0; I＆下; 4;我+ +）
   {
       D [第（i + 4  -  octetshifts）％4] =
           （（S [1]  - ;＆下; bitshift）及微米）|
           （（S [第（i + 1）％4]≥＆GT; bitsleft）及流明）;
   }
}
 

显然

I have an array of four unsigned chars. I want to treat it like a 32-bit number (assume the upper bits of the char are don't care. I only care about the lower 8-bits). Then, I want to circularly shift it by an arbitrary number of places. I've got a few different shift sizes, all determined at compile-time.

E.g.

unsigned char a[4] = {0x81, 0x1, 0x1, 0x2};
circular_left_shift(a, 1);
/* a is now { 0x2, 0x2, 0x2, 0x5 } */

Edit: To everyone wondering why I didn't mention CHAR_BIT != 8, because this is standard C. I didn't specify a platform, so why are you assuming one?

解决方案

static void rotate_left(uint8_t *d, uint8_t *s, uint8_t bits)
{
   const uint8_t octetshifts = bits / 8;
   const uint8_t bitshift = bits % 8;
   const uint8_t bitsleft = (8 - bitshift);
   const uint8_t lm = (1 << bitshift) - 1;
   const uint8_t um = ~lm;
   int i;

   for (i = 0; i < 4; i++)
   {
       d[(i + 4 - octetshifts) % 4] =
           ((s[i] << bitshift) & um) | 
           ((s[(i + 1) % 4] >> bitsleft) & lm);
   }
}   

Obviously

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