geeksforgeeks

Question

#include <stdio.h>

int main()
{
 static int i=5;
 if(--i)
  {
     main();
     printf("%d ",i);
  }
}

 

 

Comments

0 0 0 0  Should be Output

Answer 1

NOTE THAT i is static variable ===> loop must be stop at some where

printf function is called after recursion completed ====> updated value of variable is printed...

updated value = 0 at last step ===> recursion exit

0,0,0,0 as o/p

 

doubt is how many no.of zero's would print?

5--->4--->3--->2--->1

at i=1, if condition makes it 0 and evaluate as false ===> when i=1, print statement doesn't executed...

therefore only four number of zero's should print.

Comments

I mean to show how that program executes

See here main() calling 5 times and printing start when the last main is called i.e.when i=0

my question is 

at first i is printing 0

then scope of that main() is closed , I marked in previous code as................(1)

Next code.....(2) will called and scope of i is 2 here , so why not it will print value of i is 2??

 

---

mam, static values are not keep in stack, they stored in stack ===> when last update i=0, then where ever you access that variable is getting the value = 0 only.

local static variable doesn't keep a copy for each function call.

where as just normal local variable keep a copy for each function call.

---

ok understood

thanks

Answer 2

Using tree method, go top to down left to right, print value of i whenever printf(i) encountered

Comments

why not printing 0,1,2,3?scope of static is local

main()
{
    static int i=4;
    .........
    main()
    {
       static int i=3;
       ...........
       main()
       {
           static int i=2;
           ............
           main()
           {
               static int i=1;
               ............
               main()
               {
                   //i=0;return
               }
           }
           
       }
    }

 

Answer 3

You can understand more clearly using two basic concepts.

First, Static variable assign memory only once in the entire execution of a program, unlike local variable can have a reference of more than one memory after each local declaration.

Second, Visually the recursion call stored activation record in a stack.

Thing going like that:

Let the declaration of a static variable is line number 1 and accordingly other follow line number.

In each call function run only up to line number 3 and store its state in stack activation record and static value reduce its value continuously (4,3,2,1) at i=0, the condition will fail. After line number 3, the execution of other state stored in stack execute started and get the value of i=0 at each time.

There is four activation record in the stack for i=4,3,2,1 so 4 zeros print consecutively.

If this program is tail recursive, it need not store the state in stack and value print (4,3,2,1) if you call main after printf 

If you know stack and how the activation record stored in a stack then you can visualize my explanation easily. 

If anyone feels anything wrong, pls comment.

Thanks!