Return a 2D array from a function with just the array size as the arguments

Question

i am new to c++ and i am trying to create an Array from within a function with just the array size, [row][col], as the argument. I know you cannot return an array from a function in c++. I would allocate memory in these situations if i were to code in C. However, i am not sure what is the C++ way (:

I tried,

    int** get2Darray(int row, int col){
         int** randInts = new int[row][col]; //Invalid. Constant required.
         return randInts;
    }

Thanks.

EDIT: vector class does the trick, however, my program is heavily modular. I dont have access to the main program. My function should be called and a fresh array should be created and returned. Implementation should only be done inside this function

when i return a vector type, would i able to use it on a int** type? — user859385
– user859385, Commented Mar 4, 2016 at 19:22
@user859385, you are returning an array of uninitialized pointers. — blazs
– blazs, Commented Mar 4, 2016 at 19:39

blazs · Accepted Answer · 2016-03-04 18:58:06Z

3

Either change int** randInts = new int[row][col]; into int* randInts = new int[row * col]; or define int** randInts as follows:

int** randInts = new int*[row];
for (int i = 0; i < row; ++i) {
    randInts[i] = new int[col];
}

It's a C way, not a C++ way. In C++ you should use containers (e.g. a vector as suggested in the comments).

answered Mar 4, 2016 at 18:58

blazs

4,85528 silver badges38 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

The Vivandiere · Accepted Answer · 2016-03-04 19:33:13Z

3

I have shown this using vector <vector<T> >, but you can also use Eigen matrix or boost::ublas::matrix

std::vector< std::vector<int> > foo(const int rows, const int cols)
{
    std::vector< std::vector<int> > v;
    v.resize(rows);
    for (auto i = 0; i < rows; i++)
    {
        v[i].resize(cols);
    }
    return v;
}

This will return a matrix of desired dimensions, and all elements will be initialized to 0.

The std::move part invokes a behavior called move semantics. You can read up on it. It's basically a better way of returning objects across a function boundary.

For compatibility's sake, if you must have a native array, then do this :

std::vector< vector<int> > temp = foo(rows, cols);
int *v = &temp[0][0];

Now, you can implement the gymnastics to get the correct index inside of int *v. You don't need multiple indirection.

P.S. This implementation assumes that your compiler supports C++ 11.

edited Mar 4, 2016 at 19:33

answered Mar 4, 2016 at 19:03

The Vivandiere

3,2195 gold badges34 silver badges51 bronze badges

10 Comments

blazs Over a year ago

You should probably mention this is C++11.

The Vivandiere Over a year ago

@blazs agree. Thanks.

badgerr Over a year ago

Shouldn't the return type of the function be non-void?

kfsone Over a year ago

It knows to do Return Value Optimization. Scott Meyers explains it in this talk: youtube.com/watch?v=smqT9Io_bKo

kfsone Over a year ago

@FirstJens See ideone.com/OqHHzG note how "without move" directly constructs into the return value not using an intermediate.

|

Collectives™ on Stack Overflow

Return a 2D array from a function with just the array size as the arguments

2 Answers 2

Comments

10 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

10 Comments

Your Answer

Sign up or log in

Post as a guest

Related