Expert: Kaustav Neogy - 5/5/2008

Question
QUESTION: Hello i was trying to write a for loop that sums the even values from the x array and print it. eg

x[0]= 30
x[1]=  12
x[2] = 51
x[3] = 17
x[4] = 45
x[5] = 62

eg the sum for this list would be 104

i.e (30 + 12 + 62)


part b:

i wanted to this time, using the same x[] array to write a f or loop that sums the even-numbered elements(elements 0, 2, and 4) from the list. eg sum would be 126
(30 + 51 + 45.)

Here are my codes for each part but i dont know why it's not giving me the desired values. please help!!


part a

#include<stdio.h>
#include<conio.h>

#define N   6

int main()
{
   int x[] = {30, 12, 51, 17, 45, 62,};
   int i ;
   int sum = 0;
   
   for(i=0; i<N;i++)
   {
         while((x[i]%2)==0){   
   sum+= x[i];
   printf("%2d", sum);
   x[i++];
   }
  }
    getche();
    }


part b:

#include<stdio.h>
#include<conio.h>

#define N   6

int main()
{
   int x[] = {30, 12, 51, 17, 45, 62,};
   int i ;
   int sum = 0;
   
   for(i=0; i<N;i++)
   {
         while((x[i]%2)==0){   
   sum= x[0]+ x[2] + x[4];
   printf("%2d", sum);
   x[i++];
   }
  }
    getche();
    }


ANSWER: Hi Henry,

here is the updated code which should work.

part a

#include<stdio.h>
#include<conio.h>

#define N   6

int main()
{
  int x[] = {30, 12, 51, 17, 45, 62,};
  int i ;
  int sum = 0;
  
  for(i=0; i<N;i++)
  {
        if(x[i]%2==0)
         {   
         sum+= x[i];
         }  
  }
  printf("%2d", sum);

   getche();
   }


part b:

#include<stdio.h>
#include<conio.h>

#define N   6

int main()
{
  int x[] = {30, 12, 51, 17, 45, 62,};
  int i ;
  int sum = 0;
  
  for(i=0; i<N;i++)
  {
        if(i%2==0)
         {   
         sum+= x[i];
         }
  }
  printf("%2d", sum);
   getche();
   }

HTH.

---------- FOLLOW-UP ----------

QUESTION: Hi,

i wrote the code as a solution to the part b but it didn't print anything. could you tell me why? Also, explain the algorithm of my code for me to better understand arrays. i mean start with the for loop and the loop content.

#include<stdio.h>
#include<conio.h>

#define N   6

int main()
{
   int x[] = {30, 12, 51, 17, 45, 62,};
   int i ;
   int sum=0;
   
   for(i=0; i<N;i+2)
   {
   
   sum+= x[i];
   }
    printf("%2d", sum);

    getche();
    }



ANSWER: Hi,


#include<stdio.h>
#include<conio.h>

#define N   6

int main()
{
  int x[] = {30, 12, 51, 17, 45, 62,};
  int i ;
  int sum=0;

//When you do a i+2 here, the counter increases by 2 every time. So, it should sum up x[0] + x[2] + x[4] + x[6] + x[8] + x[10].  
  for(i=0; i<N;i+2)
  {
  
  sum+= x[i];
  }
   printf("%2d", sum);

   getche();
   }

HTH.

---------- FOLLOW-UP ----------

QUESTION: Hello Kaustav,

Exactly what i was expecting but i think it will only sum up x[0] + x[2] + x[4]but i think it will keep incrementing up to x[10] from what i just inferred from your explanation. therefore, how would i have done it?

Answer
Like this -

#include<stdio.h>
#include<conio.h>

#define N   6

int main()
{
 int x[] = {30, 12, 51, 17, 45, 62,};
 int i ;
 int sum = 0;
 
 for(i=0; i<N;i++)
 {
       if(i%2==0)
         {   
         sum+= x[i];
         }
 }
 printf("%2d", sum);
  getche();
  }