Getting inputs with scanf for an array - arrays

Question

Just getting to learn C better and I'm playing with arrays.
I would like to enter my phone number into an array like this:
#include <stdio.h>
int main()
{
int phoneNum[10];
for (int i = 0; i < sizeof(phoneNum); i++) {
printf("Insert digit %d of your phone number: \n", i + 1);
scanf("%d", &phoneNum[i]);
}
return 0;
}
This seems to fail as it keeps asking me for a new digit. So I tried to print the size of the array:
int phoneNum[10];
printf("%lu", sizeof(phoneNum));
which incredibly gives me the result of 40 even though I initialized it to be 10 (?).
I have three questions:
Why is the result 40 and not 10 in sizeof(phoneNum) ?
How can I add elements in an array successfully with scanf in the above manner?
If the above manner is silly, is there a better, more efficient way to do this? For example directly enter 10 digits into an array? I can of course use scanf("%d%d%d...", digit1, digit2, digit3, ...) but I would like a generalized way, using the size of the array (say I don't know it and it's passed from another function)

Answer 1

sizeof(phoneNum) returns 10 * sizeof(int). The sizeof(int) value appears to be 4 for your system.
#include <stdio.h>
int main()
{
int phoneNum[10] = {0};
const size_t size = sizeof(phoneNum) / sizeof(int);
for (int i = 0; i < size; i++) {
printf("Insert digit %d of your phone number: \n", i + 1);
scanf("%d", &phoneNum[i]);
}
for(int i = 0; i < size; ++i)
{
printf("\r\n %i \n", phoneNum[i]);
}
return 0;
}

Answer 2

sizeof(phoneNum) will return number in bytes, not length of array.
after the includes you could make a define like #define SIZE 10 and use SIZE like if it was a constant.
#include <stdio.h>
#define SIZE 10
int main()
{
int phoneNum[SIZE];
for (int i = 0; i < SIZE; i++)
{
//Do Something
}
}
Take into account the fact that strings should end with null terminated character (\0) so the size of the string have that space available.