如何在C中将位数组的元素传递给函数

Question

I would like to know how to pass an element of a bit array into a function in C language. I want to control the function execution by this "logic" signal. My idea is that some function sets or clears dedicated element of a bit array and this action changes behavior of my function but I don't know how to declare the function prototype (I don't how to say the compiler that the function expects bit array element as one of his arguments). Does it ever exist any trick how to use individual bit array elements in a C language function? Thanks for any suggestions.

Answer 1

It sounds like you are looking for some generic solution using function pointers. That is, pass a function pointer which determines the action to take. Example:

void bit_change (uint8_t*  arr, 
                 size_t    byte_index,
                 size_t    bit_index, 
                 action_t* action)
{
  arr[byte_index] = action(arr[byte_index], bit_index);
}

Where action_t is the function type determining the behavior. Now you can implement actions such as bit sets, bit clears, bit toggles, set all bits etc etc.

Full example:

#include <stdint.h>
#include <inttypes.h>
#include <stdio.h>

typedef uint8_t action_t (uint8_t data, size_t bit);

uint8_t bit_set (uint8_t data, size_t bit)
{
  return (uint8_t) (data | (1u << bit));
}

uint8_t bit_clear (uint8_t data, size_t bit)
{
  return (uint8_t) (data & ~(1u << bit));
}


void bit_change (uint8_t*  arr, 
                 size_t    byte_index,
                 size_t    bit_index, 
                 action_t* action)
{
  arr[byte_index] = action(arr[byte_index], bit_index);
}


void print_array (size_t size, const uint8_t array[size])
{
  for(size_t i=0; i<size; i++)
  {  
    printf("%.2"PRIu8 " ", array[i]);
  }
  printf("\n");
}


int main (void)
{
  uint8_t array [5] = {0};

  bit_change(array, 0, 2, bit_set);
  print_array(sizeof(array), array);

  bit_change(array, 0, 2, bit_clear);
  print_array(sizeof(array), array);

  return 0;
}