c - 如何使用C将奇数和偶数位置位存储到另一个指针？

Question

我有两个指针。一个是输入 ptr，另一个是输出..

我的输入指针包含二进制值

15.................0
 0011 0110 1111 0111 - unsigned 16 bit
 1011 0100 1011 1100 - unsigned 16 bit

首先，我想从两行中获取偶数位置位，并且必须将其填充到输出指针中。

o/p 指针将是

  0110  0110  0110  1111   (even position bits from both)..from right to left and assume 0 to 15.
  1100  1110  0101  1101   (odd position bits from both)

如何使用指针来做到这一点？我对指针很陌生

Answer 1

将代码转换为使用指针通常非常简单。

举个例子:

 uint16_t compute(uint16_t a, uint16_t b)
 {
     return ((a | b) & MY_BIT_MASK) >> (a & 2); // some arbitrary operations
 }

 uint16_t compute_p(uint16_t *a, uint16_t *b)
 {
     return ((*a | *b) & MY_BIT_MASK) >> (*a & 2); // some arbitrary operations via pointers
 }

如果您向我们展示您的实际代码，我们也许可以为您提供更多帮助。但基本上，当你有一个指向 uint16_t 的指针时，你可以通过两种方式使用它:

 uint16_t *p;

   p  <- to mean the pointers value, which is an address, you assign it to other pointers, or increment it, so it would point to the next uint16_t in memory
   *p  <-  to use it in expressions on place of a uint16_t, this way your code will each time fetch the actual 16 bit value pointed to by p
   p[i]  <-  with int i, This is basically a synonym to *(p+i) as long as i is non-negative
   *(p + i)  <-   with int i, This is also the value of 16 bit integer where p points to, or the next, or previous etc 16 bit value. In this form, negative i can be used as well
   p - q  <-  with uint16* q the difference of the pointers, in case of uint16_t * types, this basically means "how many 16 but ints I can store and addresses q, q+1, q+2, q+3 .... p-1", assuming p>q