How this program executes?

Question

#include <stdio.h>
#include <string.h>
int f(int n){
static int i=1;
return n+i--;
}
int main()
{
char ch[]="CODE";
int n = strlen(ch), i=0;
for(f(n);i=f(n);i++)
printf("%s\n",ch+i-1);

return 0;
}

Answer 1

Go to create tab in the code section tap + button and paste the code and click run.

Answer 2

cryptic codes are good for puzzles, but I hope you're not planning on using this on practical projects.

it's the same as

#include <stdio.h>
#include <string.h>

int main(){
char ch[]="CODE";
for(int i=strlen(ch);i>0;i--)
printf("%s\n",ch+i-1);
}

only more complicated.
You can put printf inside the f function to see what's going on when the program runs. The int i inside the function f is declared static, meaning it will retain its value and will not be reset between function calls. So it becomes more and more negative each time (i--) which makes returned value smaller and smaller (n+i--) every time f(n) is called.

Answer 3

maybe I've made it more complicated than necessary, but here is how you can see the values in each part of the code as the for loop runs

#include <stdio.h>
#include <string.h>
int f(int n){
static int i=1;
printf(" inside f(n) i = %d\n", i);
int returnval = n+i--;
printf(" n+i-- = %d\n", returnval);
return returnval;
}

int main(){
char ch[]="CODE";
int n = strlen(ch), i=0;

for(f(n);(i=f(n));i++){
printf(" in main() i = %d\n", i);
printf("\nch+i-1 = ch+%d = %s\n\n", i-1, ch+i-1);

}
printf("\nmain i = f(n) = %d\n", i);
return 0;
}