print an int and char with the same identifier in C - c

Question

I'm trying to do a simple code but I'm having hard time while trying to make it work.
I want to get an int from stdin between 1 and 50.
if the number is 1 then to print A
if the number is 11 then to print J
if the number is 12 then to print Q
if the number is 13 then to print K
Edit: If its none of them, then just return the number.
i tried to use %c but it wont work for numbers from 10 and above
then i managed to do it by using switch but the default part is not working for me. the only thing i managed to do is to make 50 case's for each number but that's just look horrible.
any help would be appreciated.
#include <stdio.h>
main() {
int number;
scanf("%d", &number);
char* card = NULL;
switch (number)
{
case 1:
card = "A";
break;
case 11:
card = "J";
break;
case 12:
card = "Q";
break;
case 13:
card = "K";
break;
default:
card = //Dont know what to write here//;
}
printf("%s\n", card);
return 0;
}

Answer 1

As far as I can see, this code should work:
#include <stdio.h>
#include <string.h>
#include <limits.h>
#define MAX_STRING_SIZE (sizeof(int)*CHAR_BIT/3 + 3)
int main(int argc, char **argv) {
int number = 0;
scanf("%d", &number);
char card[MAX_STRING_SIZE] = {0};
switch (number)
{
case 1:
strcpy(card, "A");
break;
case 11:
strcpy(card, "J");
break;
case 12:
strcpy(card, "Q");
break;
case 13:
strcpy(card, "K");
break;
default:
sprintf(card, "%d", number);
}
printf("%s\n", card);
return 0;
}
Note that you cannot assign strings with: card = "whatever", you must use strcpy() found in the string.h library. For the default part of your code, I assumed you were trying to convert the card input which was not 1, 11, 12, or 13. In which case, you can use sprintf() to convert the integer value to a string.
I hope this helps!

Answer 2

#define MINCARD 1
#define MAXCARD 50
const char *num2card(int n)
{
static char buf[3];
assert(n >= MINCARD && n <= MAXCARD && "Inval card no.");
switch (num) {
case 1:
return "A";
case 11:
return "J";
case 12:
return "Q";
case 13:
return "K";
default:
snprintf(buf, 3, "%d", num);
}
return buf;
}

Answer 3

Use an array of char. Each numerical "case" will correspond to an index in the array, and the element at that index will be the appropriate "card".
As for what the "default" value should be - that is a requirements question. What is the purpose of this application?

Answer 4

you can try sprintf, but you still have to add in logic for what happens if the number is larger than 50.
#include <stdio.h>
#include <stdlib.h>
int main() {
int number;
scanf("%d", &number);
char* card = malloc(8);
switch (number)
{
case 1:
card = "A";
break;
case 11:
card = "J";
break;
case 12:
card = "Q";
break;
case 13:
card = "K";
break;
default:
sprintf (card , "%i" , number );
}
printf("%s\n", card);
return 0;
}