C programming arrays assistance?

Question

Write c program where the user takes an array with n elements and finds the elements whose value is a multiple of the number 5.These elements are displays on screen along with their value like this : V[i]=a ,V[j] = b,.... My code :

#include <stdio.h>
int main ()

    {
        int n,i,sh=0;
        int v[100];
        printf ("Please write n:");
        scanf("%d",&n);
        for (i=0;i<n;i++)
        { printf ("\n Write the element %d",i);
        scanf("%d",&v[i]);
        }
        if (v[i] %5)
        printf("The element is a multiple of 5",&sh);

        return 0;
    }

It compiles perfectly,but when I run this and I write the elements,it does nothing..where am I wrong?

EDIT :

Yes,here it is :
Please write n: 4
Enter value 0: 10
Enter value 1 : 9
Enter value 2 : 20
Enter value 3:14

V[0]=10,V[2}=20

`if (v[i] %5)` is not inside the loop. it's going to always use the last value of `i`. — Jonathon Reinhart, Aug 15 '13 at 08:58
If you indent your code, you might see the problem. You do the `v[i] %5` check *outside* the loop, meaning you only do it once and you do it on an index one higher than the number of entered values. — Some programmer dude, Aug 15 '13 at 08:59
additionally `printf("The element is a multiple of 5",&sh);` is wrong — Grijesh Chauhan, Aug 15 '13 at 09:00
It is going to crash if user enters more than 100 when asked for n. — aragaer, Aug 15 '13 at 09:21

score 2 · Answer 1 · edited Aug 15 '13 at 13:58

2

This:

if (v[i] %5)
    printf("The element is a multiple of 5",&sh);

After the first loop your i is equal to n+1. You don't reset it neither check your elements in a loop.
Also your format string doesn't have any %p in it, why do you pass %sh?

Do something like:

 for (i=0;i<n;i++)
    if (v[i] %5) // not divided by 5
       // Correct prtinf() statement 
       // printf("The element v[%d] = %d is NOT multiple of 5", i, v[i]);

edited Aug 15 '13 at 13:58

Grijesh Chauhan

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answered Aug 15 '13 at 09:00

aragaer

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To whoever edited my post - not necessarily index-out-of-bounds at this point. The loop goes up to n, which can be more, less or equal than array size (which is 100). – aragaer Aug 15 '13 at 09:15

score 2 · Accepted Answer · edited May 23 '17 at 12:11

2

You have three errors in your code.

First if ( v[i] % 5 ) is outside of the loop at the end, it will just try this once for i = n+1. You will have the out-of-bound problem.
And you search a multiple of 5 so the condition should be if ( v[i] % 5 == 0 ).

After, your printf is also wrong.

printf("The element is a multiple of 5",&sh);
//                                     ^^^^

What is sh ? Why do you want to use it in printf ? You format string does not seem to want an argument.

Your code should look like this :

for ( i=0; i < n; i++ )
{
    printf( "\n Write the element %d", i );
    scanf( "%d", &v[i] );
    if ( v[i] % 5 == 0 )
        printf( "The element is a multiple of 5" );
}

EDIT :

With the out put example, maybe you should do another loop :

#include <stdio.h>

int main ()
{
    int n, i, sh = 0; // I still don't for what sh is used ...
    int v[100];

    printf ("Please write n:");
    scanf("%d",&n);

    // Get the values
    for ( i=0; i < n; i++ )
    {
        printf( "\n Write the element %d", i );
        scanf( "%d", &v[i] );
    }

    // Print the values multiple of 5
    for ( i=0; i < n; i++ )
    {
        if ( v[i] % 5 == 0 )
            printf( "V[%d]=%d\n", i, v[i] );
    }

    return 0;
}

HERE is the example working.

edited May 23 '17 at 12:11

Community

1
1

answered Aug 15 '13 at 09:06

Pierre Fourgeaud

14,290
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+ for check additional mistake in if condition – Grijesh Chauhan Aug 15 '13 at 09:09
Thank you but I have to write ALL the numbers,and then it shows which is a multiple,not one by one..thank you anyway! – user2685334 Aug 15 '13 at 09:11
Do you have an example of the output you should get ? – Pierre Fourgeaud Aug 15 '13 at 09:13
It doesnt work at all,Im sorry..now when I write n = 4,I can only enter one value.. – user2685334 Aug 15 '13 at 09:34
@user2685334 It works fine for me... Are you sure you copied this code correctly ? – Pierre Fourgeaud Aug 15 '13 at 09:40
@user2685334 I attached an example working after my example. http://ideone.com/9P236j – Pierre Fourgeaud Aug 15 '13 at 09:45

score 2 · Answer 3 · edited Aug 15 '13 at 09:24

The IF statement isn't in the loop
printf("The element is a multiple of 5",&sh); - am I missing something? (why the &sh?)
The IF statement isn't totally correct

How would I edit it:

 1|    #include <stdio.h>
 2|
 3|    int main ()        
 4|    {
 5|        int n,i,sh=0;
 6|        int v[100];
 7|        printf ("Please write n:");
 8|        scanf("%d",&n);
 9|        for (i=0;i<n;i++)
10|        { 
11|             printf ("\n Write the element %d",i);
12|             scanf("%d",&v[i]);
13|             
14|             if ((v[i] % 5) == 0)
15|                 printf("The %d. element %d is a multiple of 5", i+1, v[i]);
16|        }
17|
18|        return 0;
19|    }

Just wondering what is the purpose of variable sh? Or for what did the author wanted to use it?

score 0 · Answer 4 · answered Aug 15 '13 at 09:29

{
    int n,i,sh=0;
    int v[100];
    printf ("Please write n:");
    scanf("%d",&n);
    for (i=0;i<n;i++)
    { printf ("\n Write the element %d",i);
    scanf("%d",&v[i]);
    }

    /// now go back through the array
    //
    for( i = 0; i < n; i++ )
       if( v[i] % 5 == 0 )
           printf( "The element at %d is %d, divisible by 5\n", i, v[i] );

    return 0;
}

score 0 · Answer 5 · answered Aug 15 '13 at 14:02

int n,i,sh=0;

//array to hold the numbers
int N[100];

//count of numbers
printf ("Enter the NUMBER count :");
scanf("%d",&n);

//get the input in an array for (i=0;i

//check for the divisibility
for( i = 0; i < n; i++ )
   if( N[i] % 5 == 0 )
       printf( "The element at %d is %d, divisible by 5\n", i, N[i] );

return 0;

}

C programming arrays assistance?

5 Answers5