Getting out of scope error with dynamically allocated array - c++

Question

I am getting an out of scope error in my copy function, when I try to copy the contents of the first array into the DMA one. The try-catch block is required.
#include <iostream>
#include <cstdlib>
using namespace std;
void show( const int a[], unsigned elements );
int * copy( const int a[], unsigned els );
void die(const string & msg);
int main()
{
int arr[4] = {4, 2, 3, 6};
show(arr, 4);
int * newArr = copy(arr, 4);
}
void show( const int a[], unsigned elements )
{
for (int i = 0; i < elements; i++)
cout << a[i] << endl;
}
int * copy( const int a[], unsigned els )
{
try
{
int * newArr = new int[els];
}
catch(const bad_alloc &)
{
die("Alloc Failure");
}
for (int i = 0; i < els; i++)
newArr[i] = a[i];
return newArr;
}
void die(const string & msg)
{
cerr << "Fatal error: " << msg << endl;
exit(EXIT_FAILURE);
}

Answer 1

If you declare the variable inside the try block then it's only accessible there. You can work around this by moving the declaration outside of the block.
int *newArr;
try
{
newArr = new int[els];
}
catch(const bad_alloc &)
{
die("Alloc Failure");
}
for (int i = 0; i < els; i++)
newArr[i] = a[i];
Or by moving the rest of the code inside the try.
try
{
int *newArr = new int[els];
for (int i = 0; i < els; i++)
newArr[i] = a[i];
return newArr;
}
catch(const bad_alloc &)
{
die("Alloc Failure");
}

Answer 2

Define new array before the try, otherwise it is only defined inside the try block.

Answer 3

The whole point of exceptions is that you do not need to handle all conceivable errors right at the point where they happen, but rather in places of your choice where you are able to respond to them meaningfully. So be generous with the size of your try blocks:
int * copy(const int a[], unsigned els)
{
try
{
int * newArr = new int[els];
for (int i = 0; i < els; i++)
newArr[i] = a[i];
return newArr;
}
catch (const std::bad_alloc &)
{
die("Alloc Failure");
}
}

Answer 4

You can change your code to
int * copy( const int a[], unsigned els ) {
int * newArr = nullptr;
try {
newArr = new int[els];
}
catch(const bad_alloc &) {
die("Alloc Failure");
}
if(newArr) {
for (int i = 0; i < els; i++)
newArr[i] = a[i];
}
return newArr;
}
to overcome your problem. Just initialize newArr correctly.

Answer 5

I think the problem in your Copy-Function is, that the new Array is a local Variable. When you than return it to the caller, the pointer adresses aren't the right ones.
You might better use two arrays which you pass as Referneces into the copy-function, like this:
int& copy(int& newArray, const int& oldArray) { ... }
Here you get a discription about references in c++ if you don't know what they supposed to be: http://en.wikipedia.org/wiki/Reference_%28C%2B%2B%29