Char array can't save chars as its members. Password system using ATmega8

Question

I was work with system, that read some symbols from the specific keyboard, pass it to the ATmega8, and then pass it to the display one-by-one(this step work correctly), but if I want to show all symbols array, I discovered that the dynamic array on first position save null, on the second save empty symbol, and the subsequent characters saved correctly. I don't see any mistakes in code, so I need a help.

This code function must return the 4-elements char array of symbols reading serial from the keyboard.

char* askPass(void){
     int i;
     char key;          
     #define PS 4
     char* pass = (char*)calloc(PS, sizeof(char));

     clear:
     lcd_clear();
     lcd_gotoxy(0, 0);
     lcd_puts("Enter the password:");  
     lcd_gotoxy(0, 1);               
     lcd_puts(">>");                          
     free(pass);
     pass = (char*)calloc(0, sizeof(char));
     for (i=0;i<PS;i++) pass[i] = ''; 

     for (i=0; i<PS; i++) {
         key = '-';
         key = readKey();
         lcd_gotoxy(3+i, 1);
         if (key == 'C') {
             goto clear;
         } else if (key == '-'){
             lcd_putchar('|');
             delay_ms(10);
             lcd_gotoxy(3+i, 1);
             lcd_putchar(' ');
             delay_ms(10);
             lcd_gotoxy(3+i, 1);
             i--;
         } else {
             pass = (char*)realloc(pass, i*sizeof(char));
             *(pass+i) = key;   
             lcd_putchar(*(pass+i));
             delay_ms(20);
         }            
     }             
     /// there is an error: 
      /// serial input: 1234
      /// lcd output: !* 34! 
      /// correct output: !1234!
     lcd_gotoxy(0,2);
     lcd_puts("!");       
     for (i=0; i<PS; i++) {
         if (!(*(pass+i))) lcd_putchar('*'); 
         else lcd_putchar(*(pass+i));
     }
     lcd_puts("!");                    
     // end error block 

     return pass;
 }  // can't return correct array value

/* All tests show this:

Serial input word: abcd
Output: !* cd!
Correct output: !abcd!

*/

Answer 1

You start my allocating 4 bytes with:

char* pass = (char*)calloc(PS, sizeof(char));

Then you call free(pass) before you even use the allocated data, only to then allocates zero bytes with:

pass = (char*)calloc(0, sizeof(char));

If makes no sense to allocate only to free without using the allocation, or to allocate zero bytes. The act of allocating zero bytes is undefined, but nothing good or useful will happen by attempting to write to such an allocation as you do.

Continuously reallocation to add one byte at a time is ill-advised also. It would be better to allocate a buffer of a reasonable length to start with and if you need to extend, extend by a number of characters in on chunk.

It is unclear in any case why you are dynamically allocating and reallocating to allow input of an arbitrary number of characters that you are not using. It would be simpler and more deterministic, to simply accept the four characters in a fixed length array and discard any extraneous input.

There is no such thing as an "empty character", calloc already initialised the allocation to zero (a nul character).

I am pretty certain this code has many other issues, but it is had to determine what you are trying to do in order to advise; you would well to use a debugger.