CODEWITHSUNDEEP
carrays
Green Noob
Green Noob

Reputation: 408

Deleting an array element in C

I wrote the following program to delete an array element entered by the user.

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

void main() {
    int j, i, a[100], n, key, l;
    clrscr();
    printf("Enter the number of elements:");
    scanf("%d", &n);
    printf("\nEnter the elements:\n");
    for (i = 0; i < n; i++)
        scanf("%d", &a[i]);
    printf("\nEnter the element to delete:");
    scanf("%d", &key);
    l = n;  //Length of the array
    for (i = 0; i < l; i++) {
        if (a[i] == key) {
           for (j = i; j < l; j++)
               a[j] = a[j + 1];
           l--;    //Decreasing the length of the array
        }
    }

    printf("\nThe new array is \n");
    for (i = 0; i < l; i++)
        printf("%d ", a[i]);
     getch();
}

It works fine for most inputs but when the input is something like: 1 2 2 3 5 (here 2 repeats consecutively) and the element to be deleted is 2, the output is 1 2 3 5.

How can I modify the program such that all instances of the entered element is removed?

Upvotes: 5

Views: 36278

Answers (7)

chqrlie
chqrlie

Reputation: 145307

Your method with 2 nested for loops is too complicated. You can simple scan the array with an index i and copy all elements different from key with a different index len. The resulting array length is the final value of len.

Here is a modified version:

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

int main(void) {
    int a[100];
    int i, n, key, len;

    clrscr();
    printf("Enter the number of elements: ");
    if (scanf("%d", &n) != 1) {
        printf("invalid input\n");
        return 1;
    }
    if (n < 0 || n > 100) {
        printf("invalid number of elements\n");
        return 1;
    }
    printf("\nEnter the elements:\n");
    for (i = 0; i < n; i++) {
        if (scanf("%d", &a[i]) != 1) {
            printf("invalid input\n");
            return 1;
        }
    }
    printf("\nEnter the element to delete: ");
    if (scanf("%d", &key) != 1) {
        printf("invalid input\n");
        return 1;
    }

    for (i = len = 0; i < n; i++) {
        if (a[i] != key)
           a[len++] = a[i];
    }

    printf("\nThe new array is:\n");
    for (i = 0; i < len; i++)
        printf("%d ", a[i]);
    printf("\n");
    getch();
    return 0;
}

Notes:

  • The prototype for main without arguments is int main(void) and it is considered good style to return 0 for success.

  • always test the return value of scanf(). This prevents many bugs and undefined behavior for invalid input. It also saves a lot of time looking in the wrong places when input was just invalid.

  • avoid naming a variable l as it looks too close to 1 in many fixed pitch fonts.

  • always terminate the program output with a newline.

Upvotes: 0

Dhruv Aditya Mittal
Dhruv Aditya Mittal

Reputation: 1

#include<stdio.h>


int main(){
int size;
int array[20];
int delete_pos;
int i;

printf("Enter the Size of the Array :");
scanf("%d",&size);


for(i=0;i<=size-1;i++){                                //no of elements taken are 1 less than size of the array asked.
    printf("\nEnter the element[%d] :",i+1);
    scanf("%d",&array[i]);
}

printf("\nEnter the Position of the array to be deleted :");
scanf("%d",&delete_pos);


for(i=delete_pos-1;i<=size;i++){                    //every element of the array is replaced by array on next position.
    array[i]=array[i+1];}

size=size-1;                                       // Reducing the size of the array as one element is deleted.
printf("Your new array is \n");
for(i=0;i<=size-1;i++){                            //printing the entire new array.
    printf("%d ",array[i]);

}
printf("\n\n");
return 0;
}

Upvotes: 0

FrankTan
FrankTan

Reputation: 1686

use a new array.

int array[l];
int k=0;
for(i=0;i<l;i++)
{
 if(a[i]!=key)
 {
  array[k]=a[i];
  k++;
 }

}

Upvotes: 0

ugoren
ugoren

Reputation: 16449

Change "if" to "while":


    for(i=0;i<l;i++)
    {
        while (i<l && a[i]==key)
        {
            for(j=i;j<l;j++)
                a[j]=a[j+1];
            l--;    //Decreasing the length of the array
        }
    }

Upvotes: 1

user97370
user97370

Reputation:

If you don't care about the order of the elements in the array, you can move the last element of the array into the newly formed gap (cunningly reducing the length of the array by one). This can be vastly more efficient than shunting the elements down: in computer science term this makes deleting an element O(1) rather than O(N).

a[i] = a[--l];

If your i index is looping over the array, you'll want to loop over this element again:

a[i--] = a[--l];

For example, to remove all elements '3' from an array of length 'l':

for (i = 0; i < l; ++i) {
    if (a[i] == 3) {
        a[i--] = a[--l];
    }
}

If you do care about the order of the elements in the array, it's most efficient to use memmove rather than move elements by hand. It's designed to be used where the source and destination memory overlap.

memmove(a + i, a + i + 1, sizeof(a[0]) * (l - i - 1));

Upvotes: 3

pmg
pmg

Reputation: 108986

Other posters have given you 2 solutions ... I think understanding why it happens is good too :)

Let's take your example 1, 2, 2, 3, 5 and follow the code line by line

i = 0;             /* first time through the loop; i "points" to 1 */
if (a[i] == 2) ... /* nope; next loop */
i = 1;
if (a[1] == 2) ... /* yes! let's go inside the if */
                   /* move all elements back
                   ** and "decrease" array length */
                   /* array is now 1, 2, 3, 5 */
                   /* next loop */
i = 2;
if (a[i] == 2) ... /* nope; OH! Wait ...
                   ** a[1] is the new 2 and it wasn't checked */

Upvotes: 2

Shadow
Shadow

Reputation: 6287

After l-- put i-- too as shown below

if(a[i]==key)
  {
   for(j=i;j<l;j++)
    a[j]=a[j+1];
   l--;    //Decreasing the length of the array
   i--;    //check again from same index i
  }

Upvotes: 5

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