C Programming - Functions - Discussion

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.

You are wrong yash we can't use two return statemt in a common statement.

Yes we can use two or more return statements in a common function but if there are no conditions then it returns threw the first return statement. The others do not execute.

One thng is that why it is returning zero?

Can we two retun statement in a common function defination?

Please give me some sugetion to make my "c" strong?

Vivek explained nice.

#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?

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)

@Hagesh

Can you explain this in clear way?

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

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.