C Programming - Functions - Discussion
Discussion Forum : Functions - Find Output of Program (Q.No. 6)
6.
What will be the output of the program?
#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;
}
Discussion:
52 comments Page 5 of 6.
Chatrapathi said:
1 decade ago
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.
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.
Madhu said:
1 decade ago
@Teja. You make a small mistake.
In first executing loop you have written that sumdig (n) =0;.
Why it is zero. ?
It is not zero. It will execute sumdig (n) function again.
In first executing loop you have written that sumdig (n) =0;.
Why it is zero. ?
It is not zero. It will execute sumdig (n) function again.
Satyanarayana said:
1 decade ago
Very nice vivek.
Kantasiva said:
1 decade ago
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;
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;
Piyush Sinha said:
1 decade ago
Actually u people are missing a small thing... the output is 3+2+1+0. as the return 0 is in else part which comes when if(n!=0) becomes false and then it return s. I hope everything will be clear now. :)
8246 said:
1 decade ago
First d=3 and n=12.
Next d=2 and n=1.
Next d=1 and n=0.
Therefore answer is 3+2+1=6.
Next d=2 and n=1.
Next d=1 and n=0.
Therefore answer is 3+2+1=6.
Yasin said:
1 decade ago
@ Auto Variable Initialization
S value is initialized with Garbage value.
But, this value is over written with s=d+sumdig(N);
S value is initialized with Garbage value.
But, this value is over written with s=d+sumdig(N);
Taher Ali said:
1 decade ago
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.
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.
Srinivas kumar said:
1 decade ago
d%10 gives the last number of a given number i.e., for example for number 123 last number is 3 and n/10 gives the except last number it gives whole number i.e., for example for number last number is 3 and remaining are 12.
Like that you can solve the code.
Like that you can solve the code.
Vijaya said:
1 decade ago
If we have two return stmts without any condition then the first return statement is executed the next one is ignored. In the above program why the output 0 and 0 can anyone explain me.
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