Deleting an array element in C

Question

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

#include 
#include 

void main() {
    int j, i,          


        

Answer 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;
}

Answer 2

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.

Answer 3

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++;
 }

}

Answer 4

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
        }
    }

Answer 5

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
  }

Answer 6

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 */