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:
51 comments Page 1 of 6.
Ammu said:
8 years ago
step 1: d=123%10==3
n=123/10==12
s=3+sumdig(12)since n==12.
step 2: d=12%10==2
n=12/10==1
s=5(since 3+2)+sumdig(1).
step 3: d=1%10==1
n=n/10==0
s=5+1==6
so s==6.
n=123/10==12
s=3+sumdig(12)since n==12.
step 2: d=12%10==2
n=12/10==1
s=5(since 3+2)+sumdig(1).
step 3: d=1%10==1
n=n/10==0
s=5+1==6
so s==6.
(7)
Divya gawande said:
7 years ago
Why in output 6, 6 is two times taken? Please explain in detail.
(2)
Debjani said:
6 years ago
Why the value of b is 6?
(2)
Noel said:
7 years ago
I think the sticking point for me and others is once you have d = 1%10.
If it were division, the result would be 0. However, with MODULAR MATH, 1%10 is 1.
If it were division, the result would be 0. However, with MODULAR MATH, 1%10 is 1.
(1)
Shalom chauhan said:
1 decade ago
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
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
(1)
Ram said:
7 years ago
Thanks @Ammu.
(1)
Vivek said:
1 decade ago
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)
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)
(1)
N DAMODHAR said:
9 years ago
Can anyone explain this the program.
I don't understand the step in s = d+sumdig(n);
I don't understand the step in s = d+sumdig(n);
Shahid said:
8 years ago
@Taher Ali.
Awesome solution. It works for (- and *).
Awesome solution. It works for (- and *).
Divya said:
1 decade ago
Can someone explain, when n=0 after n=n/1;if (n!=0) condition becomes false.
So, it should return 0. How can it even reach the statement returns?
So, it should return 0. How can it even reach the statement returns?
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