C Programming - Functions - Discussion

First of all thanks sree for explaining.

Now if we can understand how 0 is printed first then we can understand the whole program.

BRIEFED ANSWER IS IN STEP 3 MUST READ

Firstly if you can figure out this program you can understand above one.

FUNCTION A(INT N)
{
IF(N>0)
{
FUNCTION B() // calls function b
PRINT STATEMENT
}
}

1. Function b is called

2. Function b fails.

3. FUNCTION B RETURNS TO IF STRUCTURE AFTER FAILING AND THEN PRINT STATEMENT EXECUTED and 0 is printed.

4. Then further functions called.

Can someone please explain this in the most simple manner?

fun(3) //upside fun() running
if(3>0)
fun(2) //upside fun() running
if(2>0)
fun(1) //upside fun() running
if(1>0)
fun(0) //upside fun() running
if(0>0) false
so return; //upside fun() finished
"print == 0"

fun(-1) //Now downside fun() running
if(-1>0) false
so return; //Downside fun() finished
"print == 1" //BCoz Inside the fun(1) block fun(0)

if(0>0) false
so return; //Downside fun() finished
"print == 2" //BCoz Inside the fun(2) block fun(1)

if(1>0)
fun(0) false
so return;
"print == 0" //BCoz Here last n=0
fun(-1)
if(-1>0) false
so return; //Downside fun() finished

OUTPUT == 0,1,2,0

Thank you kavita.

Thanks Sree

Its correct answer Thanks Sree

Thanks Ritesh. You are excellent.

void fun(int n) is a function
and in this function we again calling fun(--n)

Means it is a recursive function

Here the condition if(n>0) again check

This works as a loop till condition is false

That means till i=0

Then it prints 0

After print again we are calling fun(--n)

Now 2 loops have to execute so

Finally it prints 120.

I will try to explain:

First note that for these things:
1. fun(0) will not execute.
2. fun(1) will only print 0 i.e. --n.

Now follow the tree

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

fun(1)----> print 0

So output would be 0 1 2 0

Thank you rithesh. Good explanation. :).