Program 280: Print Sum of Even Numbers in Array using Recursion

Program 280: Print Sum of Even Numbers in Array using Recursion

#include<stdio.h>
void SumOfEven(int a[],int num,int sum);
main()
{
    int i,a[100],num,sum=0;
    printf("Enter number of Array Elements\n");
   scanf("%d",&num);
   printf("Enter Array Elements\n");
   for(i=0;i<num;i++)
   {
    scanf("%d",&a[i]);
    }
   SumOfEven(a,num-1,sum); 
}

void SumOfEven(int a[],int num,int sum)
{
 
 if(num>=0)
 {
  if((a[num])%2==0)
  {
   sum+=(a[num]); 
  }
  SumOfEven(a,num-1,sum);
 }
 else
 {
    printf("Sum=%d\n",sum); 
  return;
 }
}

Explanation:

1. This program starts with initializing :
   • i→ Helping variable
   • a[100[→To sore numbers entered by user
   • num→ Total number of elements n array

   • void SumOfEven(int a[],int num,int sum);→Declaring function

   • sum→ To store sum of the even numbers

2.  printf("Enter number of Array Elements\n");
     scanf("%d",&num);
      printf("Enter Array Elements\n");
      for(i=0;i<num;i++)
      {
       scanf("%d",&a[i]);
       }

   Taking numbers from user and storing in a

3. SumOfEven(a,num-1,sum);

   Calling SumOfEven Function where we are passing array a ,total number of elements as num-1 because we have to calculate from 0 to n-1 as array starts with 0 so if we give 7 elements in array then recursion will execute 7 times i.e from 0-6, and sum which is initialized to zero

4. void SumOfEven(int a[],int num,int sum)
   {
    
    if(num>=0)
    {
     if((a[num])%2==0)
     {
      sum+=(a[num]); 
     }
     SumOfEven(a,num-1,sum);
    }
    else
    {
       printf("Sum=%d\n",sum); 
     return;
    }
   }

   Now lets take 4 elements in array 30,35,40,45 so sum should be 30+40=70.We passed this array to SumOfEven so num=(4-1)=3 and sum=0 this is our initial data
5. Now
   

   1. if(num>=0)
       {
        if((a[num])%2==0)
        {
         sum+=(a[num]); 
        }
        SumOfEven(a,num-1,sum);
       }

      Since num=3 which is >=0 so if part will execute.
      • a[num]=a[3]=45
      • a[num]%2=45%2!=0. So sum+=a[num] will not execute
      • Finally sum=0
      • Now we are calling SumOfEven(a,num-1,sum) again
      • So we are passing SumOfEven(a,3-1,0) ---->SumOfEven(a,2,0)
   2. The same process in step 1 continue again

   3. if(num>=0)
       {
        if((a[num])%2==0)
        {
         sum+=(a[num]); 
        }
        SumOfEven(a,num-1,sum);
       }

      Since num=2 which is >=0 so if part will execute.
      • a[num]=a[2]=40
      • a[num]%2=40%2==0. So sum+=a[num] will execute
      • sum=0+40=40
      • Finally sum=40 in recursion 2
      • Now we are calling SumOfEven(a,num-1,sum) again
      • So we are passing SumOfEven(a,2-1,40) ---->SumOfEven(a,1,40)
   4. Since num=1 which is >=0 so if part will execute.
      • a[num]=a[1]=35
      • a[num]%2=35%2!=0. So sum+=a[num] will not execute
      • Finally sum=40 in recursion 3 and no change is there
      • Now we are calling SumOfEven(a,num-1,sum) again
      • So we are passing SumOfEven(a,1-1,40) ---->SumOfEven(a,0,40)
   5. Since num=0 which is >=0 so if part will execute.
      • a[num]=a[0]=30
      • a[num]%2=30%2==0. So sum+=a[num] will execute
      • sum=40+30=70
      • Finally sum=70 in recursion 4
      • Now we are calling SumOfEven(a,num-1,sum) again
      • So we are passing SumOfEven(a,0-1,70) ---->SumOfEven(a,-1,70)
   6. Since num=-1 which is  not >=0 so else part will execute.

   7.  else
       {
          printf("Sum=%d\n",sum); 
        return;
       }

      • So here sum will be printed which is 70
      • And then it returns back to main()

Output: