C calling free on not allocated memory

Why this program doesn’t crash, is it undefined behaviour ?

int main()
{
   char* c;
   if (c) {
        printf("called free\n");
        free (c);
   }
   else {
        printf("not called free\n");
   }

   printf("not crashed\n");
   c = strdup("someString");

   if (c) {
        printf("called free\n"); 
        free(c);
        c = NULL;  // why is this needed for last if (c) 
   }
   
   if (c) {
        free(c);
   }
   printf("still not crashed\n");

    return 0;
}

1. Is any possibility that this program will ever crash ?
2. Should I always make such a if (c) verification before I call free on char* ?
3. why without c=NULL free is called ?

>Solution :

Is any possibility that this program will ever crash ?

1. Yes, there is. Your code invokes undefined behaviour;

behavior, upon use of a nonportable or erroneous program construct or
of erroneous data, for which this International Standard imposes no
requirements.

Anything can happen, and that includes a crash.

Should I always make such a if (c) verification before I call free on
char* ?

No, a call to free() with a NULL pointer constant is a NOP, i.e. no operation is performed. So it’s unnecessary.

why without c=NULL free is called ?

Because the free() function doesn’t set the original pointer to NULL. c is still pointing to the same memory location, but it is invalid for c to access it anymore. Without setting c to NULL, a subsequent call to free() would be freeing already freed memory, which also invokes undefined behaviour.