How do i pass an array function without using pointers

Question

I have been asked in an interview how do you pass an array to a function without using any pointers but it seems to be impossible or there is way to do this?

Answer 1

simply pass the location of base element and then accept it as 'int a[]'. Here's an example:-

    main()
    {
        int a[]={0,1,2,3,4,5,6,7,8,9};
        display(a);
    }
    display(int a[])
    {
        int i;
        for(i=0;i<10;i++) printf("%d ",a[i]);
    }

Answer 2

Put the array into a structure:

#include <stdio.h>
typedef struct
{
  int Array[10];
} ArrayStruct;

void printArray(ArrayStruct a)
{
  int i;
  for (i = 0; i < 10; i++)
    printf("%d\n", a.Array[i]);
}

int main(void)
{
  ArrayStruct a;
  int i;
  for (i = 0; i < 10; i++)
    a.Array[i] = i * i;
  printArray(a);
  return 0;
}

Answer 3

How about varargs? See man stdarg. This is how printf() accepts multiple arguments.

Answer 4

If i say directly then it is not possible...!

but you can do this is by some other indirect way

1> pack all array in one structure & pass structure by pass by value

2> pass each element of array by variable argument in function