Appending a dynamic array and doubling its size upon completion - c++

Question

Create a dynamic array of int with a initial space of 4. Write a function ‘append’ that appends a given value to this array. At any stage, if this function finds the array full it automatically doubles the size of array to accommodate this new value. Also write a function to display all the elements of this array. Write a main to test all these functions.
I made an attempt to solve the above question as given below. However, I am unable to get the correct code. Please help me
#include<iostream>
using namespace std;
void append(int*& array, int val, int n, int r)
{
int i,p;
int *array1;
for (i=r;i<n;i++)
array[i] = val;
if(i==n)
{
p = 2*n;
array1 = new int [p];
}
for(int j =0; j<r/2; j++)
array1[j]= array[j];
append(array1, val, p, p/2);
}
int main()
{
int q,p=0,val, n = 4;
int n1 = p/2;
int *array = new int[n];
while(1)
{
cout<<"Enter 0 to end and 1 to continue";
cin>>q;
while(q!=0)
{
cin>>val;
append(array,val,n,n1);
}
}
return 0;
}
I need to solve this without using "Classes". How shall I do it?

Answer 1

Your function needs to do the following:
1) Be able to check if the current append call will result in an out-of-bounds write attempt. So you need something like (and give variables explanatory names like this) this as the first line in your function:
if (array_size < item_count) {
//double size of array
}
To double the size of the array, you have to make a new array with twice the size, copy all the items over from the old array, DELETE the old array, null the old array's pointer, and somehow update the array_size variable (return to main is one option, a static int counter in the function itself is another). You may have to return a pointer to the new array to main, as well. Or maybe you can just re-address the old pointer to the new array AFTER using that pointer to delete the old array. This is all about avoiding memory leaks. So, try to come up with a method declaration like:
int append(int* arrPtr, int value, int array_size, int item_count)
This particular approach means main is getting sent back the array size as an int after each append. So you need something in main like:
array_size = append(array, val, array_size, item_count);
The tricky part will be when you make the new array:
array_size = 2 * array_size;
int* temp = new int[array_size]
//copy everything over from old array to new using arrPtr and temp
for (int i = 0; i < array_size / 2; i++) {
temp[i] = arrPtr[i]
}
//delete the contents of the old array:
delete[] arrPtr;
//pointer gymnastics to redirect arrPtr to the new array:
arrPtr = temp;
temp = nullptr;
//okay, now you are back out of the conditional and can use item_count to insert the
//value into the array (doubled or not)
arrPtr[item_count] = value;
//don't forget to return array_size, and remember main must track item_count as well.
return array_size;
That's the general gist of it. This is not a complete answer, but should give you enough to work with. Basically, most of your code has to be rewritten, and the above is not a complete solution. Good luck.

Answer 2

After taking cue from Double size of dynamic array I have solved it.
#include<iostream>
using namespace std;
void add_element(int* &array, int &size)
{int count = 0;
while(1)
{
int number;
cout << "What number do you want to add? " << endl;
cin >> number;
if (count == size)
{
int newSize = size * 2;
int *newArr = new int[newSize];
for (int i = 0; i < count; ++i)
{
newArr[i] = array[i];
}
delete[] array;
array = newArr;
size = newSize;
}
array[count] = number;
++count;
int k;
cout<<"Do u want to end, then press 0";
cin>>k;
if(k==0) break;
}
for(int g = 0; g<count; g++)
cout<<array[g]<<'\t';
}
int main()
{
int i,j,k,size;
cin>>size;
int* array = new int [size];
add_element(array, size);
}