trouble returning an array of a structure - c

Question

I'm a newbie and I have a problem with array of type structure. I don't really know where to specify that the array is a pointer and where to not specify it. So whatever the combination I do with pointers in declaration and in calling of function involving this array of structure, it doesn't compile. Do you have any idea why ?
here is the code in question :
typedef struct{
char letter[7];
} player;
player init_player(){
player tab[2];
char letter_player1[4]={' ','a','b','c'};
char letter_player2[4]={' ','x','y','z'};
strcpy(tab[1].letter,letter_player1);
strcpy(tab[2].letter,letter_player2);
return *tab;
}
void displaying(player tab){
int position=1;
printf("%c\n",*tab[1].letter[position]);
}
int main(){
player tab=init_player(tab);
displaying(tab);
}
Thank you in advance !

Answer 1

How your code is wrong:
/* required #include (or function prototypes) are not written */
typedef struct{
char letter[7];
} player;
player init_player(){ /* the return type is not for returning arrays */
player tab[2];
/* no terminating null characters are placed, so cannot be used with strcpy() */
char letter_player1[4]={' ','a','b','c'};
char letter_player2[4]={' ','x','y','z'};
strcpy(tab[1].letter,letter_player1);
strcpy(tab[2].letter,letter_player2); /* only tab[0] and tab[1] are available, tab[2] is out-of-bounds */
return *tab; /* not the array but only the first element of the array is returned */
}
void displaying(player tab){ /* the argument type is not for getting arrays */
int position=1;
/* even if tab were an array(pointer), dereferencing is done via [] operator, so you don't need * here */
printf("%c\n",*tab[1].letter[position]);
}
int main(){
player tab=init_player(tab); /* the variable type is not for dealing with arrays */
displaying(tab);
}
How to fix:
/* add #include */
#include <stdio.h> /* for printf() */
#include <stdlib.h> /* for malloc(), free() and exit() */
#include <string.h> /* for strcpy() */
typedef struct{
char letter[7];
} player;
/* change return type to return a pointer */
player* init_player(){
/* you cannot use non-static local array, so instead dynamically allocate an array here */
player* tab = malloc(sizeof(*tab) * 2);
/* add terminating null characters (increase array size, and they will be automatically added) */
char letter_player1[5]={' ','a','b','c'};
char letter_player2[5]={' ','x','y','z'};
if (tab == NULL) exit(1); /* for in case the allocation failed */
/* use tab[0] and tab[1] instead of tab[1] and tab[2] */
strcpy(tab[0].letter,letter_player1);
strcpy(tab[1].letter,letter_player2);
return tab; /* return the (pointer to the first element of) array */
}
/* change argument type to get a pointer */
void displaying(player* tab){
int position=1;
/* remove extra * (indirection operator) */
printf("%c\n",tab[1].letter[position]);
}
int main(){
/* change variable type to store a pointer */
player* tab=init_player(tab);
displaying(tab);
free(tab); /* the array is dynamically allocated, so clean-up it */
}
You don't need the argument tab passed to init_player(),
but it is harmless because init_player() can accept any arguments and the arguments are not used.