C Programming - Functions - Discussion

@All.

Although It happens simultaneously, Let's understand the concept as per pre-decrement.
There is --n so first n becomes n-1 and then it goes into the recursion function....

if(n > 0)
{
fun(--n);
printf("%d,", n);
fun(--n);
}

Explained :

f(3){
n = 2

f(2){
n = 1

f(1){
n = 0

f(0){
return to f(1)
{
as per value of n of above f(1)
print(0) -> 0

return to f(2)
{
as per value of n of above f(2)
print(1) -> 1

now, n=0

f(0){
return to f(3){
as per value of n of above f(3)
print(2) -> 2

now, n=1
f(1){
n=0
f(0){
return to f(1){

as per recent value of f(1){
print(0) -> 0

now, f(0){
return

return from f(3).

So, the final answer series is 0, 1, 2, 0.

If you use stack concept then question contain stack key.

The Correct answer for this question is, no output. Because if the condition is true then the loop will be created. If the condition is false(n=0)it do not execute if condition. Then how control goes to printf()?

I have a doubt when the functions are called recursively each time n value decremented by 1 hence finally it comes 0 so now n value in fun (0) is 0 so now n=0 then how it prints 1 2 in fun (1) fun (2) because here also n must be 0 how it happens? Please tell me.

It is a stack concept.

When n=3 then it becomes 3,2,1,0 as per condition, these values are stored in the stack
0 1 2 3.

When the condition becomes false then automatically values are poped. Here top value is 0 or last inserted so 0 is first poped and execution jumps to print function and display 0 and same for all.

At last output will be;
0 1 2 0

// typedef int (*pf) (int, int);
// int proc(pf, int, int);

The above two statement has no effects , because they are not used in main or sub function.

Function calls are stored in stack, that is LAST IN FIRST OUT.

1.fun(a=3)---calling sub function--->fun(n=3).
2.fun(3)->if(3>0)true{fun(2)-->step 3,print(n),fun(--n)//active}
3.fun(2)->if(2>0)true{fun(1)-->step 4,print(n),fun(--n)//active}
4.fun(1)->if(1>0)true{fun(0)-->step 5,print(n),fun(--n)//active}
5.fun(0)->if(0>0)false;
6.From step 4,{fun(n=0)//finished,PRINT(n=0),fun(-1)-->step 7}
7.fun(-1)->if(-1>0)false;
8.From step 3,{fun(n=1)//finished,PRINT(n=1),fun(0)-->step 9}
9.fun(0)->if(0>0)false;
10.From step 2,{fun(n=2)//finished,PRINT(n=2),fun(1)-->step 11}
11.fun(1)->if(1>0)true{fun(0)->step 12,print(n),fun(--n)//acti}
12.fun(0)->if(0>0)false;
13.From 11,{fun(n=0)//finished,PRINT(0),fun(-1)-->step 14}
14.fun(-1)->if(-1>0)false;

// all active calls are called and removed from stack.
// here 'n' is a local variable for each function.
// The printing statements are capitalized

PRINT(n=0),PRINT(n=1),PRINT(n=2),PRINT(n=0).

Step 1: fun (3) ->3>0.

1.1: fun (2) ->2>0, still lines left to be processed.

1.2: fun (1) ->1>0, still lines left to be processed.

1.3: fun (0) ->0>0, condition fails comes to the next line print the current value of n (i.e. ) n=0; prints 0, next fun (-1) comes out of the current 1. 3 loop.

Now again going back to unprocessed lines in step 1.2,

1.2: Unprocessed line here in this call starts from printf. For this loop n value is 1; prints 1, & comes to next line.

fun (--n) ->fun (0) 0>0 fails. 1.2 loop ends here.

Now again going back to unprocessed lines in step 1.1,

1.1: Unprocessed line here in this call starts from printf. For this loop n value is 2; prints 2, & comes to next line.

fun (--n) ->fun (1) ; once again calling the function fun (1).

1st line - decremented the value to 0;

2nd line prints the value; (prints 0,).

3rd line - fun (-1) totally terminates everything.

Final o/p : 0, 1, 2, 0.

Thanks all for giving the explanation.

@Raj.

Your explanation is the best. Thanks.

How can I recognize a function is a recursion?

Please explain me.

According to me, It is;

fun(3)->fun(2) -> fun(1) -> fun(0)
print 2 print 1 print 0 finish
fun(1)-> fun(0) fun(-1)
print 0 finish finish
finish