How to properly set pointers to an array with chars in it

Question

I hope some of you may help me with a basic understanding of pointers assigned to an array with chars in it (specifically letters). My problem is that part of my code when the user is supposed to input letters, 'D' or 'U' doesn't work and I don't know exactly where the problem is. I was trying to set it without pointers but that made my program more complicated and I'm also trying to finally learn that goddamn thing.

#include <stdio.h>
#include <stdlib.h>

int main()
{
   int number;

   scanf("%d", &number);
   char *pz;
   char save[number];
  

   for (int i = 0; i < number; i++)
   {
       pz = &save[i];
       scanf("%s", pz+i);
   }

Answer 1

It seems that instead of this

   for (int i = 0; i < number; i++)
   {
       pz = &save[i];
       scanf("%s", pz+i);
   }

you mean

   for (int i = 0; i < number; i++)
   {
       pz = save + i;
       scanf( " %c", pz );
   }

Pay attention to the blank in the format string " %c".

The pointer pz already points to the desired position in the array

pz = save + i; 

So using the expression pz+i in the call of scanf does not make a sense.

And instead of the conversion specifier %s you have to use the conversion specifier %c.

After inputting the character array you can output it like

printf( "%.*s\n", number, save );

Or you can output the array in a loop like for example

   for ( pz = save; pz != save + number; ++pz )
   {
       putchar( *pz );
   }
   putchar( '\n' );