C Programming - Functions - Discussion

In the above program, they mention printf only once after performing sumdig functions. Therefore, the program prints the output values of sumdig values only.

What to do with return 0 statement?

Why the value of b is 6?

@Deepak and @Debjani.

As we are calling the same function with same parameter in the main.
That value is assigned to 'b'.

So same operation takes place as 'a'.
And then b also get 6.

@Hakesh.
Return 0 belongs to else part of the function.
If n=0 then value 0 is returned .

In the program, the value n is not equal to 0.
So, if the condition is executed and there is no work with else part.
So the value of s is returned.

@Vivek.

You explained it in a best way. Thank you very much.

Thanks for explaining the answer.

Hi
See this code
if(n!=0)
{
1. d = n%10;
2. n = n/10;
3. s = d+sumdig(n);
}

In 1. the value of d became 3 //remainder
In 2. the value of n became 12 //quotient
In 3. the value of s became 3+0 //initially sumdig(n)=0
Again
In 1. the value of d became 2
In 1. the value of d became 1
In 2. the value of n became 0 // 1/10=0
In 3. the value of s became 1+5
The condition false the control come to end of function

a=sumdig(123)

n=123 which n!= 0, So n will enter into the loop.

Now in the loop the values of d, n, s are 3(123%10), 12(123/10), 3+sumdig(12) [which will recurcive call].

So finally we will get the value for s is 6(3+2+1).

a=6;

Same as we find the value of b. b=6;

printf("%d%d",a,b); will print 6 and 6.

Both function works as same so the output is 6 and 6..