Question

Can someone help me understand this step by step (C++)

Array Shifter and Array Reversal Write a function that accepts an array of doubles and the array's size as arguments. The function should create a new array that is one element larger than the argument array. The first element of the new array should be set to 0. Element 0 of the argument array should be copied to element 1 of the new array, element 1 of the argument array should be copied to element 2 of the new array, and so forth. The function should return a pointer to the new array. Write a second function that accepts an array of doubles and the array's size as arguments. The function should create a new array that contains all of the original array's elements in reverse order. Last of all, have a main function that demonstrates the correct working of these two functions.

Answer 1

#include <iostream>
using namespace std;

// Function to insert 0in arr at position 0
double* insert(double arr[],
int n)
{
int i,pos=0;

// increase the size by 1
n++;

// shift elements forward
for (i = n; i >= pos; i--)
arr[i] = arr[i - 1];

// insert x at pos
arr[pos] =0;
return arr;
}

double* reverse(double arr[], int n)
{
double aux[n];

for (int i = 0; i < n; i++) {
aux[n - 1 - i] = arr[i];
}

for (int i = 0; i < n; i++) {
arr[i] = aux[i];
}
return arr;
}

// Driver Code
int main()
{
double arr[10] ={1.2,2.3,3.6,5.6,9.3};
int i, x, n = 5;

for (i = 0; i < n; i++)
cout << arr[i] << " ";
cout << endl;

insert(arr,n);
cout << "array with 0 at 1st position"<<endl;
// print the updated array
for (i = 0; i < n + 1; i++)
cout << arr[i] << " ";
cout << endl;
double arr2[10] ={1.2,2.3,3.6,5.6,9.3};
reverse(arr2,n);
cout << "reversed array:"<<endl;
for (i = 0; i < n ; i++)
cout << arr2[i] << " ";
cout << endl;

return 0;
}

// THANKING YOU...