c programming

Question

#include <stdio.h>
char *fun()
{
    static char arr[1024];
    return arr;
}
 
int main()
{
    char *str = "geeksforgeeks";
    strcpy(fun(), str);
    str = fun();
    strcpy(str, "geeksquiz");
    printf("%s", fun());
    return 0;
}

What would be the output?

Answer 1

char *str = "geeksforgeeks";
//a RO memory is created and "geeksforgeeks"
//stored there and address assigned to str
strcpy(fun(), str);
//fun returns the address of arr and content
//of str copied there.
str = fun();
//value of str changes to that of arr
strcpy(str, "geeksquiz");
//"geeksquiz" copied to memory pointed to by atr(arr)
printf("%s", fun());
//Content from memory pointed to by arr printed

Comments

str = fun();
what is meaning of this line?

---

see fun() is returning starting address of arr. So this address will be stored in str

Answer 2

arr is a static array so it's content will remain in the memory even after execution of the function fun()

str = fun(); ---------------- becoz of this statement str contains address of arr

strcpy(str, "geeksquiz"); -------------------- now "geeksquiz" will be stored in str which is pointing to arr

printf("%s", fun()); ------------------- so if we try to print arr we will get "geeksquiz"