How to copy values from an array into a new one?

Question

I've been trying to figure this out off and on for a week now and I keep running into problems.

My objective:

Write a function that allocates memory for an integer array. The function takes as an argument an integer pointer, the size of the array, and newSize to be allocated. The function returns a pointer to the allocated buffer. When the function is first called, the size will be zero and a new array will be created. If the function is called when the array size is greater than zero, a new array will be created and the contents of the old array will be copied into the new array. Your instructor has provided arrayBuilder.cpp as starter code for this programming challenge. In addition, Lab9_1.exe is the executable for this application which you can test.

The code:

#include <iostream>
using namespace std;

int * arrayBuilder(int * arr, int size, int newSize);
void showArray(int * arr, int size);

int main()
{
int * theArray = 0;
int i;

cout << "This program demonstrates an array builder function." << endl << endl;

// create the initial array.  The initial size is zero and the requested size is 5.
theArray = arrayBuilder(theArray, 0, 5);

// show the array before values are added
cout << "theArray after first call to builder: " << endl;
showArray(theArray, 5);

// add some values to the array
for(int i = 0; i < 5; i++)
{
    theArray[i] = i + 100;
}

// show the array with added values
cout << endl << "Some values stored in the array: " << endl;
showArray(theArray, 5);

// expand the size of the array.  size is not the original size.  newSize
// must be greater than size.
theArray = arrayBuilder(theArray, 5, 10);

// show the new array with the new size
cout << endl << "The new array: " << endl;
showArray(theArray, 10);

cout << endl;

delete [] theArray; // be sure to do this a1t the end of your program!

system("pause");

return 0;
}

/*
FUNCTION: arrayBuilder
INPUTS Pointer to an array.  Size of the array. If size is zero, arr can be    NULL.
      Size of the new array.
OUTPUTS:  Returns a pointer to allocated memory.  If newSize is greater than size,
      an array of newSize is allocated and the old array is copied into the new
      array. Memory pointed to by the old array is deleted.  All new elements
      are initialized to zero.
*/


int * arrayBuilder(int * arr, int size, int newSize)
{
// TODO: Your code goes here


return NULL; // default return value.  No memory allocated!
}

/*
FUNCTION: showArray
INPUTS: Pointer to an array.  Size of the array. If size is zero, arr can be  NULL.
OUTPUTS:  Prints the contents of the array to the console.
*/


void showArray(int * arr, int size)
{
cout << "arr = ";

for(int i = 0; i < size; i++)
{
    cout << arr[i] << "  ";
}

cout << endl;

}

My struggles: I cannot figure out how to switch "arr" and a temporary array's values.

int * arrayBuilder(int * arr, int size, int newSize)
{
// TODO: Your code goes here
    int * temp = new int [newSize];

for (int i = size; i < newSize; i++)
{
        *arr = *temp;
        temp++;
}

return NULL; // default return value.  No memory allocated!
}

another attempt while searching for answers:

int * arrayBuilder(int * arr, int size, int newSize)
{
// TODO: Your code goes here
int * temp = new int [newSize];
memcpy (temp, arr, size *sizeof(int));
// HINT: Design the function before writing it.
delete[]  arr;

for (int i = size; i < newSize; i++)
{
    temp[i] = i;
}

return NULL; // default return value.  No memory allocated!
}

Basically my end goal is to have the answer look like this:

This program demonstrates an array builder function.

theArray after first call to the builder:
arr = 0 0 0 0 0

some values stored in the array:
arr = 100 101 102 103 104

the new array:
arr = 100 101 102 103 104 0 0 0 0 0

PROGRESS!! Its not crashing anymore :-) This is where I'm at now:

This program demonstrates an array builder function.

theArray after first call to builder:
arr = -842150451  0  0  0  0

Some values stored in the array:
arr = 100  101  102  103  104

The new array:
arr = -842150451  -842150451  -842150451  -842150451  -842150451  -842150451  -8
42150451  -842150451  -842150451  -842150451

Press any key to continue . . .

I'll keep tinkering and let everyone know if I hit a wall! Thanks again guys!

OKAY! finally got it to display properly:

This program demonstrates an array builder function.

theArray after first call to the builder:
arr = 0 0 0 0 0

some values stored in the array:
arr = 100 101 102 103 104

the new array:
arr = 100 101 102 103 104 0 0 0 0 0

This is what I did. I feel like I may have cheated in the second part when i put 0 values in for "temp". It was my understanding that i was going to take the data from the previous array and put it into the new one, and instead I just remade it. (So it will only work with this particular set of values [only 0's]). Is there a different way I can code the second part so it works universally with whatever values are thrown at it???

int * arrayBuilder(int * arr, int size, int newSize)
{
int i = size;
int * temp = new int [newSize];
// What if the size is 0?
if (size <= 0)
{
    while (i < newSize)
    {
        temp[i] = 0;
        i++;
    }
}
// Assuming the size _isn't_ 0
else 
{
// "a new array will be created"  (good)

for (i = 0; i < newSize; i++)
{
    // The contents of the "old" array (arr) will be
    // copied into the "new" array (temp)
    while (i < size)
    {
        temp[i] = arr[i];
        i++;
    }
    while (i >= size && i < newSize)
    {
        temp[i] = 0;
        i++;
    }
    // as a hint, you can address the elements in 
    // both arrays using the [] operator:
    // arr[i]
    // temp[i]

}
}

// "The function returns a pointer to the allocated buffer."
// So, NULL is wrong, what buffer did you allocate?
return temp; // default return value.  No memory allocated!
}

Answer 1

Since you put forth some effort.

Write a function that allocates memory for an integer array.

The prototype for this function was provided for you:

int * arrayBuilder(int * arr, int size, int newSize);

The function takes as an argument an integer pointer, the size of the array, and newSize to be allocated. The function returns a pointer to the allocated buffer.

This says nothing about doing anything with the "old" (passed in) array, so we should assume it needs to be left alone.

When the function is first called, the size will be zero and a new array will be created.

The above text is meaningless given the context. Feel free to tell your instructor I said so. If the size is zero, how do you know how many elements to allocate?

If the function is called when the array size is greater than zero, a new array will be created and the contents of the old array will be copied into the new array.

OK, now the guts of what needs to be done (you're so close)

int * arrayBuilder(int * arr, int size, int newSize)
{
    // What if the size is 0?

    // Assuming the size _isn't_ 0
    // "a new array will be created"  (good)
    int * temp = new int [newSize];

    for (int i = size; i < newSize; i++)
    {
        // The contents of the "old" array (arr) will be
        // copied into the "new" array (temp)

        // as a hint, you can address the elements in 
        // both arrays using the [] operator:
        // arr[i]
        // temp[i]

        // something is wrong here...
        *arr = *temp;

        // you definitely _don't_ want to do this
        temp++;
    }

    // "The function returns a pointer to the allocated buffer."
    // So, NULL is wrong, what buffer did you allocate?
    return NULL; // default return value.  No memory allocated!
}

Answer 2

You already got the answer here:

memcpy (temp, arr, size *sizeof(int));

but you are making several other mistakes after that. Primarily, you need to return temp ; not return NULL ;

But also you don't need the loop after the delete arr[] ;

Also don't delete arr[] if size is zero.