c++, minify dynamic array by delete [] x[i]

Question

I have a 2D array with size 4 (4 rows). To make the array size 2 (2 rows), can I use following? (for our hw assignment details are not specified and the code should be suitable with common c++ standarts)

I am removing second half of the array.

const int newSize = flightsArraySize/2;
for(int i = newSize-1; i < flightsArraySize-1; i++)
   delete [] flights[i];

Or do I have to recreate flights array with size 2?

Answer 1

Supposing that you have created a 2D array using new like this:

int **arr = new int*[rows];
for(int i=0; i<rows; ++i)
    arr[i] = new int[cols];

Then to resize it you'd have to do something like:

int newRows = rows/2;

// Create a new array for the right number of rows.
int **newArr = new int*[newRows];

// Copy the rows we're keeping across.
for(int i=0; i<newRows; ++i)
    newArr[i] = arr[i];

// Delete the rows we no longer need.
for(int i=newRows; i<rows; ++i)
    delete[] arr[i];

// Delete the old array.
delete[] arr;

// Replace the old array and row count with the new ones.
arr = newArr;
rows = newRows;

But seriously, this is all so much easier if you just use vector:

std::vector<std::vector<int>> v(rows);
for(int i=0; i<rows; ++i)
    v[i].resize(cols);
v.resize(v.size()/2);