C Programming - Functions - Discussion

step1:-> call to function sumdig with parameter 123,now n=123
step2:-> function starts
step3:-> declaration of s & d and contains null value
step4:-> 123!=0 condition true, loop starts
step5:-> d=123%10, d=3
step6:-> n=123/10, n=12
step7:-> s=d+sumdig(12),call to function sumdig with parameter 12, n=12 & s=3+sumdig(12)
step8:->step1 with value of n=12
step8:->finally s=3+2+1,6
step9:->now (n!=0) condition false come out of loop and return value 6
step10:->same execution for b=sumdig(123)
by shalom chauhan

step1:-> call to function sumdig with parameter 123,now n=123
step2:-> function starts
step3:-> declaration of s & d and contains null value
step4:-> 123!=0 condition true, loop starts
step5:-> d=123%10, d=3
step6:-> n=123/10, n=12
step7:-> s=d+sumdig(12),call to function sumdig with parameter 12, n=12 & s=3+sumdig(12)
step8:->step1 with value of n=12
step8:->finally s=3+2+1,6
step9:->now (n!=0) condition false come out of loop and return value 6
step10:->same execution for b=sumdig(123)

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

#include<stdio.h>
int sumdig(int);
int main()
{
int a, b;
a = sumdig(123);
b = sumdig(123);
printf("%d, %d\n", a, b);
return 0;
}
int sumdig(int n)
{
int s, d;
if(n!=0)
{
d = n%10;
n = n/10;
s = d+sumdig(n);
}
else
return 0;
return s;
}

Can anyone pls give the flow of control of this program?

sumdig(123) -->returns the sum of the individual digits of a number( here 123 )

The Logic here is simply taking each and every digit at a time and adding of all those digits.

i) first of all modulo division of a number gives remainder(i.e., 3) and normal division is storing in a same 'n'

Repeating the above process until we get zero in a while loop condition.

Initially n,d contains garbage value,

step-1-> N(start)=123 D=3 N=12 S=3+step2
step-2-> N(start)=12 D=2 N=1 S=2+step3
step-3-> N(start)=1 D=1 N=0 s=1+step4
step-4-> If condition is violated return 0.

So, while returning.

step-4=0
s=1+step4 becomes 1 which is step3
s=2+step3 becomes 3 which is step2
s=3+step2 becomes 6 which is returned to main.

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.

See @Vijaya. There are two return statements..but one for else..that means,

int sumdig(int n)
{
int s, d;
if(n!=0)
{
d = n%10;
n = n/10;
s = d+sumdig(n);
}
else
{
return 0;
}
return s;
}

And make sure that its if condition not while loop.

Here in this program this function (sumdig(123)) will be continue .it is nothing but the recursion function here we are using .
So
if(n!=0)
{
here n=123, 2nd loop n=12,3rd loop n=1, 4th loop =0 then the condition break;
s=d+sumdig(n) , lst time .3+2,5+1=6;

First :

a = sumdig(123)==>> sumdigt function.

d = n%10==> remainder 3 i.e d=3.
Then it go's to n= n/10==> n= 12.

s=d+sumdigit()==> sumdigit again called does the same here.
n=12 now.

d=n%10 => d=2.
n= n/10 => n=1.

Finally s=3+2+1=6.