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Topic: C

Expert: Kaustav Neogy
Date: 5/5/2008
Subject: about this code pls

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();
  }

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