How does one print out all existing integers in an int array?

Question

The code below generates the following error:

• format specifies type 'int' but the argument has type 'int *' [-Werror,-Wformat]
  printf("%i", array);

include

include

int main(void) {
    int array[6];
    for (int i = 0; i < 6; i++) {
        array[i] = get_int();
     }
printf("%i", array);

}

Answer 1

If you want to print each value in the array, you need to use a loop, just as you did for populating the array.

for (int i = 0; i < 6; i++)
{
    printf("%i", array[i]);
}