Segmentation fault when there is just one value as input - c

Question

My code seems to be working fine, but I get a Segmentation fault when there is just one value as input. It should print a square shape based on a number as character input.
test case : ["2", "2"]
"oo\noo\n"
test case: ["", ""]
""
test case : ["2"]
SIGSEGV (signal 11)
#include <string.h>
#include <stdlib.h>
void my_square(int *x, int *y)
{
int i , j;
if (*x == 0 || *y == 0) {
printf("");
}
else{
for(i = 0; i < *x; i++){
for(j = 0; j < *y; j++){
if(*x<=2 && j == 0){
printf("o");
}else if(*x<=2 && j == 1){
printf("o\n");
}else if(*y<=2 && i == 0){
printf("o");
}else if(*y<=2 && i == 1){
printf("o\n");
}else{
//printf(" i: %d, j: %d ", i, j);
if(i == 0 && j == 0 || i == *y-1 && j == 0 || i == 0 && j == *x-1 || i == *y-1 && j == *x-1){
printf("o");
}
if(i >= 1 && j == 0 && i != *y-1) {
printf("|");
}
if(i >= 1 && j == *x-1 && i != *y-1) {
printf("|");
}
if(i == 0 && j >= 1 && j != *y-1|| i == *x-1 && j >= 1 && j != *y-1){
printf("-");
}
if(i >= 1 && j >= 1 && i < *x-1 && j < *y-1){
printf(" ");
}
if(j == *x-1){
printf("\n");
}
}
//printf("> %d, %d", i, j);
}
}
}
}
int main(int ac, char **av)
{
int x = atoi(av[1]);
int y = atoi(av[2]);
my_square(&x, &y);
return 0;
}```

Answer 1

You should always check ac before accessing av, otherwise it may lead to undefined behaviour (and cause a segmentation fault).
That's how you could do it (the first value is always the program file name):
int main(int ac, char **av)
{
int x, y;
if (ac <= 3)
{
x = atoi(av[1]);
y = x; // if there's only one argument, we use it for both x and y
if (ac == 3)
{
y = atoi(av[2]);
}
my_square(&x, &y);
}
return 0;
}