如何使用C将奇数和偶数位置位存储到另一个指针？

Question

I have two pointer. one is input ptr and another one is for output..

My input pointer contains the binary values as

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

First i want to take even poistion bits from both rows and have to fill this into the output pointer.

o/p pointer will be

  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)

how to do it by using pointers?? I am very new to pointers

Answer 1

Transforming your code to use pointers is pretty simple usually.

Take the example:

 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
 }

If you show us your actual code, we can perhaps help you more. But basically, when you have a pointer to uint16_t, you can use it in two ways:

 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

Answer 2

As far as I know, you cannot do what you desire with pointers in C. This is because, as far as I know, you cannot make pointers to point bits in C, they can only point to bytes.

One solution that I can come up with is to use bit masks. By masking individual bits, shifting them into the positions that you want, and assigning them into a new blank variable could achieve what you want. Here, examine the following:

#define bitmask(_x_) (1 << (_x_))
#include <stdio.h>
#include <inttypes.h>

int main( ){

    uint16_t in1 = 14071;   // 0011 0110 1111 0111
    uint16_t in2 = 46268;   // 1011 0100 1011 1100

    uint16_t out1 = 0;
    uint16_t out2 = 0;

    for ( int i = 0; i < 8; i++ ) {
        out1 |= ( in1 & bitmask( i * 2 ) ) >> i;
        out1 |= ( ( in2 & bitmask( i * 2 ) ) >> i ) << 8;

        out2 |= ( in1 & bitmask( i * 2 + 1 ) ) >> ( i + 1 );
        out2 |= ( ( in2 & bitmask( i * 2 + 1 ) ) >> ( i + 1 ) ) << 8;
    }

    printf( "out1: %u\nout2: %u", out1, out2 );

    return 0;
}

The output is:

out1: 26223
out2: 52829

Which have the form that you want in base 2.

Edit: Alternate version with the same output, this may be somewhat more reader-friendly, I have no idea about the performance aspect:

#define bitMASK(maskee,bit) (((maskee) >> (bit)) & 1)
#include <stdio.h>
#include <inttypes.h>

int main( ){

    uint16_t in1 = 14071;   // 0011 0110 1111 0111
    uint16_t in2 = 46268;   // 1011 0100 1011 1100

    uint16_t out1 = 0;
    uint16_t out2 = 0;

    for ( int i = 0; i < 8; i++ ) {
        out1 |= bitMASK( in1, 2 * i ) << i;
        out1 |= bitMASK( in2, 2 * i ) << ( 8 + i );

        out2 |= bitMASK( in1, 2 * i + 1 ) << i;
        out2 |= bitMASK( in2, 2 * i + 1 ) << ( 8 + i );
    }

    printf( "out1: %u\nout2: %u", out1, out2 );

    return 0;
}