Copy 2D array using memcpy?

Question

So I want to copy the contents of a 2D array to another array of the exact same type. Here is how the array is created:

 GridUnit** newGrid;
 newGrid = new G         


        

Answer 1

If you actually had a 2-D array, that memcpy call would work. But you don't, you have width separate discontiguous 1-D arrays, collected in an array of pointers.

It is possible to dynamically allocate a contiguous block where row and column count both vary at runtime, and preserve the two-subscript access. You'll have to change the allocation code as follows:

GridUnit** newGrid;
newGrid = new GridUnit*[width];
newGrid[0] = new GridUnit[width * height];
for (int i = 1; i < width; i++)
    newGrid[i] = newGrid[i-1] + height;

Deallocation becomes simpler:

delete[] newGrid[0];
delete[] newGrid;

There's no delete[] for newGrid[i] with i > 0 because they don't have their own blocks, they just point into the single large block. Because everything is contiguous, you can think of newGrid[0] either as a pointer to the first row (height elements), or the entire 2-D array (width * height elements).

And you can then access all the data as a single contiguous block:

memcpy(newGrid[0], oldGrid[0], height * width * sizeof newGrid[0][0]);

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Of course, one shouldn't use raw pointers for memory ownership. A smart pointer will ensure the memory is properly deleted even with exceptional flow control. It would look like this:

std::unique_ptr<GridUnit[]> newData;
std::unique_ptr<GridUnit*[]> newGrid;
// before C++14 use // newData.reset(new GridUnit[width * height]);
newData = std::make_unique<GridUnit[]>(width * height);
// before C++14 use // newGrid.reset(new GridUnit*[width]);
newGrid = std::make_unique<GridUnit*[]>(width);
for (int i = 0; i < width; i++)
    newGrid[i] = &newData[i * height];