C++ Pointer Arrays

Question

Code

#include "stdafx.h"
#include <iostream>

void someFunc(double* pDoubleArray, int length)
{
    double* pNewDoubleArray = new double[length];

    for(int i = 0; i < length; i++)
    {
        pNewDoubleArray[i] = i * 3 + 2;
    }

    pDoubleArray = pNewDoubleArray;
}
int main()
{
    double dbls[] = { 1, 2, 3, 4, 5 };

    int length = sizeof dbls / sizeof dbls[0];

    std::cout << "Before..." << std::endl;

    for(int i = 0; i < length; i++)
    {
        std::cout << dbls[i] << ", ";
    }

    std::cout << std::endl;

    someFunc(dbls, length);

    std::cout << "After..." << std::endl;

    for(int i = 0; i < length; i++)
    {
        std::cout << dbls[i] << ", ";
    }

    std::cout << std::endl;

    while(true){ }

    return 0;
}

Output

Before...
1, 2, 3, 4, 5,
After...
1, 2, 3, 4, 5,

Here's what I am trying to do: 1. Create an array and fill it with some values 2. Pass that array as a pointer to a function that will create a new array and reassign the one that was passed in to the newly created array 3. Print out the changes

I am not seeing any changes though, and I do not know why.

Answer 1

The interface of your function someFunc is wrong. It should require the reference of a pointer's address (or the pointer to a pointer) so that you can return the address of your new array. Otherwise, you are simply modifying a local value.

void someFunc(double*& pDoubleArray, int length)
{
  double* pNewDoubleArray = new double[length];

  for(int i = 0; i < length; i++)
  {
    pNewDoubleArray[i] = i * 3 + 2;
  }

  pDoubleArray = pNewDoubleArray;
}

Your calling main function should then pass a value which can be modified:

int main()
{
  double dbls[] = { 1, 2, 3, 4, 5 };
  double* pArray = dbls;
  // ...

  someFunc(pArray, length);
  // ...

  for(int i = 0; i < length; i++)
  {
    std::cout << pArray[i] << ", ";
  }
  // ...
}

Answer 2

Ignoring the memory leak issue that results:

void someFunc(double* & pDoubleArray, int length)
// pass by reference ^^^ the pointer

Answer 3

The line     pDoubleArray = pNewDoubleArray; assigns the local copy of the pointer

Either pass the pointer by reference, pass a pointer to it, or return the new value

My preference would be to return the new value, bit that's a style issue.

Answer 4

It is unclear why you are passing the old array to a function which does not use it.

If you are changing individual values, then there is no point in creating a new array instance. If not, then simply create a new array and return it.

So, either change the original array:

void someFunc(double* pDoubleArray, int length)
{
    for(int i = 0; i < length; i++)
    {
        pDoubleArray[i] = i * 3 + 2;
    }
}

Or return the new array from the function:

// this indicates that the returned value is
// actually a new instance
double* getNewArray(double* pDoubleArray, int length)
{
    double* pNewDoubleArray = new double[length];

    for(int i = 0; i < length; i++)
    {
        pNewDoubleArray[i] = i * 3 + 2;
    }

    return pNewDoubleArray;
}

An alternative is to pass the input array by reference, but that complicates releasing unused instances.

[Edit]

To clarify this last case:

void someFunc(double** pDoubleArray, int length)
{
    double* pNewDoubleArray = new double[length];

    for(int i = 0; i < length; i++)
    {
        pNewDoubleArray[i] = i * 3 + 2;
    }

    *pDoubleArray = pNewDoubleArray;
}

void main()
{
    double dbls[] = { 1, 2, 3, 4, 5 };
    double* pArray = dbls;

    // this will change what pArray
    // points to
    someFunc(&pArray, 5);

    return 0;
}

As I've commented before, the latter approach will lead to memory leaks if pArray points to a heap allocated array before calling someFunc.