Passing an array from a struct as an argument to a function - arrays

Question

#include <cs50.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
typedef struct
{
int votes;
}
candidate;
void array_function(int arr[]);
int main(void)
{
candidate candidates[3];
candidates[0].votes = 5;
candidates[1].votes = 3;
candidates[2].votes= 1;
print_array_function(candidates.votes);
}
void print_array_function(int arr[])
{
for (int i = 0; i < 3; i++)
{
printf("%i\n", arr[i]);
}
}
I'm trying to run this code which declares a struct, feeds values into it and tries to pass an array inside the struct to a function. However, on doing so I get the following error:
test.c:22:30: error: member reference base type 'candidate [3]' is not a structure or union
array_function(candidates.votes);
How do I pass this structs array into the function?

Answer 1

Just declare the function like
void array_function( const candidate arr[], size_t n );
and call like
print_array_function( candidates, 3 );
The function can be defined the following way
void array_function( const candidate arr[], size_t n )
{
for ( size_t i = 0; i < n; i++ )
{
printf( "%i\n", arr[i].votes );
}
}

Answer 2

Here, you have an array of structure candidate which each element contains a single int called votes (certainly vote would be a better name for it). Maybe the best approach to do what you want is to create a function print_candidate_array_vote as :
void print_candidate_array_vote(candidate *candidates, unsigned int n_candidates)
{
for (int i = 0; i < n_candidates; i++) {
printf("%d\n", candidates[i].votes);
}
}

Answer 3

Try the code below, have made some changes
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
typedef struct candidate
{
int votes;
}
candidate;
void array_function(int arr[]);
int main(void)
{
candidate candidates[3];
candidates[0].votes = 5;
candidates[1].votes = 3;
candidates[2].votes= 1;
print_array_function(candidates);
}
void print_array_function(candidate arr[])
{
for (int i = 0; i < 3; i++)
{
printf("%i\n", arr[i]);
}
}