How do i decrease the size of array? - c

Question

#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
void main()
{
int i=0,pos;
int *ptr;
clrscr();
ptr=(int *)malloc(10*sizeof(int));
printf("Enter 10 Elements:\n");
while(i<10)
{
scanf("%d",&ptr[i]);
i++;
}
printf("Current Array:\n");
for(i=0;i<10;i++)
{
printf("%d\t",ptr[i]);
}
printf("Enter the position to be deleted:");
scanf("%d",&pos);
for(i=pos;i<10;i++)
{
ptr[pos-1]=ptr[pos];
pos++;
}
for(i=0;i<10;i++)
{
printf("%d\t",ptr[i]);
}
getch();
}
Array i enter : 1 2 3 4 5 6 7 8 9 0
Suppose 4th position has been deleted
Array after deletion: 1 2 3 5 6 7 8 9 0 0
How can i make the program to print only 9 integers but without decresing the times the loop run.
i.e. without using
for(i=0;i<9;i++)

Answer 1

C doesn't allow you to do that. So, alternative is to dynamically allocate an array, put what you want in it, then when you get the size you want, transfer that data to a newly allocated array of the appropriate size and delete the old one.

Answer 2

Use a variable to keep track of how many elements of the array are being used.
When you start:
unsigned num_elements = 10;
int *ptr = malloc(num_elements * sizeof(int));
When you delete, do the following as well:
--num_elements;
When you print:
for (unsigned i=0; i<num_elements; ++i) {
...
}
Since you print the array in multiple places in the code, it would be nice to create a small function to do it.
void print_array(int *a, unsigned size) {
if (size == 0) {
printf("[empty]\n");
return;
}
printf("%d", a[0]);
for (unsigned i=1; i<size; ++i) {
printf("\t%d", a[i]);
}
printf("\n");
}
print_array(ptr, num_elements);

Answer 3

Your questions refers to 2 issues here:
Deleting an element of an dynamically allocated array
Printing values of an dynamically allocated array without adjusting for loop parameters
1)
for(i=pos;i<10;i++)
{
ptr[pos-1]=ptr[pos];
pos++;
}
Does not decrease the array at all - you just overwrite the particalur position with another element but you still have memory allocated for 10 int
One way to reach that would be using realloc after you moved the values by to the left beginning at the target element:
for (i = pos; i < array_size - 1; i++)
{
ptr[i] = ptr[i+1];
}
ptr = realloc(ptr, --array_size)
To 2)
Simple answer: This wont be possible. When you decrease the size of your array and dont adjust the loop's paramter you will access memory which doesnt belong to the array anymore.
Of course you dont want to adjust every for-loop by hand but you can simplify life if you always remember the array's size, preferably in immediate reference to it within a struct:
typedef struct
{
size_t size;
int *array;
} my_array_t;
If you ensure that size is properly adjusted everytime you allocate the array's memory (functions are your friend), you can always print the values with:
my_array_t a;
a.size = 10;
a.array = malloc(a.size * sizeof(int));
for (i=0; i < a.size; i++)
// ...