Pure Pointer notation in C - c

Question

I have this coding assignment where I have to use pure pointer notation only. I am pretty much finished with it but I just realized that I used an array. I am not allowed to do so, unless I change it into a pointer somehow. That's where I am slightly stuck.
This is my code.
#include <stdio.h>
#include <stdlib.h>
/* Function Prototypes */
int main();
void s1(char *random);
void s2(char *s2_input, int index);
void strfilter(char *random, char *s2_input, char replacement);
int main()
{
for(;;)
{
int s1_index = 41;
char s1_random[s1_index];
s1(s1_random);
printf("\ns1 = ");
puts(s1_random);
printf("s2 = ");
int s2_index = 21;
char s2_input[s2_index];
s2(s2_input, s2_index);
if(s2_input[1] == '\0')
{
printf("Size too small");
exit(0);
}
if(s2_input[21] != '\0' )
{
printf("Size too big");
exit(0);
}
printf("ch = ");
int replacement = getchar();
if(replacement == EOF)
break;
while(getchar() != '\n');
printf("\n");
strfilter(s1_random, s2_input, replacement);
printf("\ns1 filtered = ");
puts(s1_random);
printf("Do you wish to run again? Yes(Y), No(N) ");
int run = getchar();
// or include ctype.h and do:
// run == EOF || toupper(run) == 'N'
if(run == EOF || run == 'N' || run == 'n')
break;
while(getchar() != '\n');
}
}
void s1(char *random)
{
int limit = 0;
char characters;
while((characters = (('A' + (rand() % 26))))) /* random generator */
{
if(limit == 41)
{
*(random + 41 - 1) = '\0';
break;
}
*(random + limit) = characters;
limit++;
}
}
void s2(char *s2_input, int index)
{
char array[21] = "123456789012345678901"; /* populated array to make sure no random memory is made */
char input;
int count = 0;
int check = 0;
while((input = getchar() ))
{
if(input == '\n')
{
*(s2_input + count) = '\0';
break;
}
else if(input < 65 || input > 90)
{
printf("invalid input");
exit(0);
}
*(s2_input + count) = input;
count++;
}
index = count;
}
void strfilter(char *random, char *s2_input, char replacement) /* replacement function */
{
while(*s2_input)
{
char *temp = random;
while(*temp)
{
if(*temp == *s2_input)
*temp = replacement;
temp++;
}
s2_input++;
}
}
My issue is this part I am not sure how to edit this to not include an array, and still have it output the program in the same way.
if(s2_input[1] == '\0')
{
printf("Size too small");
exit(0);
}
if(s2_input[21] != '\0' )
{
printf("Size too big");
exit(0);
}
I tried to take the address of the array at a certain point, and then dereference it with a pointer, however that is still using a array. Which is what I am trying to avoid. Any help would be greatly appreciated!

Answer 1

s2_input[i] can be written as *(s2_input+i) where i is some index.

Answer 2

if ((s2_input[1]) == '\0')
is equivalent to:
if (*(s2 + 1) == '\0')
Which means to dereference the value at s2 (which is the location of the zeroth[0] element), and add one to it.
The same could be done for any other location.

Answer 3

Pointer notation and what is often referred to as Indexed notation (which uses the [ ] subscript operator) are entirely equivalent. Either notion provides the pointer address plus an offset from that pointer address. See C11 Standard - 6.5.2.1 Array subscripting That being array[offset] or *(array + offset)1
For example accessing the first element using *array is shorthand for *(array + 0) which is simply array[0] in indexed notation. The 0 being the offset (in elements of that type) from the original pointer address. (type controls pointer arithmetic)
So array[10] is simply *(array + 10). If array is type char, array[10] is 10-bytes after array address. If array is type int (where an int is 4-bytes), then array[10] is 40-bytes after array address (10-int).
For a 2D array, the notation for arr2d[1][2] would simply be *(arr2d[1] + 2) which expanded further is simply *(*(arr2d + 1) + 2).
So generally array[i] is *(array + i) and arr2d[i][j] is *(*(arr2d + i) + j).
footnotes:
It follows that array[offset] is equivalent to *(array + offset) is equivalent to offset[array].