A little problem with functions and arrays - c

Question

I'm doing a program that check if 5 numbers that the user insert are even or odd, then they will be stored into an array and finally these values will be printed out on screen. In order to do this i've divided this program in two functions just to understand how the functions and the arrays works together, but it doesn't print the values that i've putted in. Why?
int check_even_and_odd(int number, int list[]){
printf("Insert the numbers\n");
scanf("%d", &list[number]);
if (number % 2 == 0) {
printf("Even\n");
}
else{
printf("Odd\n");
}
return 0;
}
int main () {
int k;
int i = 0;
int list2[5] = {0};
while (i < 5) {
i++;
k = check_even_and_odd(i, &list2[i]);
}
i = 0;
while (i < 5) {
i++;
printf("\n%d\n", list2[i]);
}
return 0;
}
Edit: Now that the main issue is gone, I want to add a little improvement to this little project. I want that the program tells to me how many Even or Odd number are in the array, but i don't know how to do it. I was thinking about adding 2 counters into the if statement (one for the even number and one for the odd numbers) but once i do this i don't know how to continue.
The program with the counters is this:
void check_even_and_odd(int number, int list[]){
int even = 0;
int odd = 0;
printf("Insert the numbers\n");
scanf("%d", &list[number]);
if (number % 2 == 0) {
even++;
}
else{
odd++;
}
printf("Even numbers are: %d\n", even);
printf("Odd numbers are: %d\n", odd);
}
int main () {
int i = 0;
int list2[5] = {0};
while (i < 5) {
i++;
check_even_and_odd(i, list2);
}
i = 0;
while (i < 5) {
i++;
printf("\n%d\n", list2[i]);
}
return 0;
}
Obviously it isn't complete, but as i have already said, i don't know how to continue

Answer 1

Your function expects an array argument but you are passing the address of individual elements of the array, so it won't work properly, you'll just need to use the correct argument:
k = check_even_and_odd(i, list2);
Quibble: k is never used so you don't really need it. You can just make your function void and remove the variable:
void check_even_and_odd(int number, int list[]){
printf("Insert the numbers\n");
scanf("%d", &list[number]);
if (number % 2 == 0){
printf("Even\n");
}
else{
printf("Odd\n");
}
}
int main(){
int i = 0;
int list2[5] = {0};
while (i < 5){
i++;
check_even_and_odd(i, list2);
}
i = 0;
while (i < 5){
i++;
printf("\n%d\n", list2[i]);
}
return 0;
}

Answer 2

Your fault is in line scanf("%d", &list[number]); just need to change it to scanf("%d", &list); but i think you are miss understanding whole array and pointer logic. You can't pass list as argument and if you do that, The compiler will changed it to pointer automatically. So if you want to tell a function about your list you just have to pass it your list address in memory (pointer). So you should do it like:
#include <stdio.h>
int How_Many_Odd = 0;
int How_Many_Even = 0;
void Add_To_List(int Number, int *ListIndex){
printf(
"Number %d is %s\n",
Number,
(Number % 2 == 0)? "Even": "Odd" // check if is odd or even
);
if(Number % 2 == 0)
How_Many_Even++;
else
How_Many_Odd++;
// changing value of pointer ListIndex to Number
*ListIndex = Number;
}
int main(){
// first creating integer array with size of 5
int List[5];
for(int i=0; i < 5; i++){
// waiting for user to enter number
int value;
scanf("%d", &value);
// changing value of index 0 to 3
Add_To_List(value, &List[i]);
}
// showing how many odds and how many evens
printf("%d numbers are even and %d numbers are odd\n", How_Many_Even, How_Many_Odd);
// you can show every index value too
for(int i=0; i < 5; i++)
printf("value of index %d is %d\n", i, List[i]);
return 0;
}
I recommend you to learn about pointer that will fix your issues

Answer 3

Here is your code fixed:
#include <stdio.h>
void check_even_and_odd(int number, int list[], unsigned int *even_count)
{
printf("Insert the numbers\n");
scanf("%d", &list[number]);
if (number % 2 == 0) {
printf("Even\n");
*even_count += 1;
}
else
{
printf("Odd\n");
}
}
int main()
{
unsigned int even_count;
int i = 0;
int list2[5] = {0};
while (i < 5)
{
check_even_and_odd(i, list2, &even_count);
i++;
}
i = 0;
while (i < 5)
{
printf("\n%d\n", list2[i]);
i++;
}
printf("There are %d even numbers and %d odd ones.\n", even_count, 5 - even_count);
return 0;
}
Basically, your main problem is in passing the list to the function in k = check_even_and_odd(i, &list2[i]);, as you should be passing the entire list, not a specific number in the list.

Answer 4

regarding:
if (number % 2 == 0) {
This is checking the passed in parameter rather than the value entered by the user. Suggest:
...
if( (list[number] % 2) == 0 )
{
printf( "%s\n", "number is even" );
....