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11-Array Expander: Write a function that accepts an int array and the array s si

ID: 3546797 • Letter: 1

Question

11-Array Expander:

Write a function that accepts an int array and the array s size as arguments. The

function should create a new array that is twice the size of the argument array. The

function should copy the contents of the argument array to the new array, and initialize

the unused elements of the second array with 0. The function should return a

pointer to the new array.


12-Element Shifter

Write a function that accepts an int array and the array s size as arguments. The

function should create a new array that is one element larger than the argument array.

The rst 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.


I need the code in C++ please.

Explanation / Answer

// PROGRAM 1 SOLUTION

#include<iostream>
using namespace std;

int* array_expander(int array[],int size)
{
int* pointer_to_new_array = new int[2*size];
for(int i=0; i<2*size; i++)
{
if(i<size)
pointer_to_new_array[i] = array[i];
else
pointer_to_new_array[i]= 0;
}
return pointer_to_new_array;
}

int main()
{
int array[]={1,2,3,4,5};
int *new_array = array_expander(array,5);
cout <<"new array contents are ";
for(int i=0; i<10; i++)
cout << new_array[i] << " " ;
cout << endl;
return 0;
}


// PROGRAM 2 SOLUTION


#include<iostream>
using namespace std;

int* element_shifter(int array[],int size)
{
int* pointer_to_new_array = new int[size+1];
pointer_to_new_array[0]=0;
for(int i=1; i<=size; i++)
{
pointer_to_new_array[i] = array[i-1];
}
return pointer_to_new_array;
}

int main()
{
int array[]={1,2,3,4,5};
int *new_array = element_shifter(array,5);
cout <<"new array contents are ";
for(int i=0; i<6; i++)
cout << new_array[i] << " " ;
cout << endl;
return 0;
}

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