How to initialize a two dimensional array using initializer_lists?

Question

I've coded this constructor to initialize my two-dimensional array using initializer_lists.

using namespace std;

class TwoArray{
    int** array;

public:
    TwoArray(initializer_list<initializer_list<int>> list_){
        const size_t row_size = list_.size();
        const size_t column_size = list_.begin()->size();
        
        array = new int[row_size][column_size]{};
    }
};

But, this code shows this error:

main.cpp: In constructor ‘TwoArray::TwoArray(std::initializer_list<std::initializer_list<int> >)’:
main.cpp:19:48: error: array size in new-expression must be constant
         array = new int[row_size][column_size]{};
                                                ^
main.cpp:19:48: error: the value of ‘column_size’ is not usable in a constant expression
main.cpp:13:22: note: ‘column_size’ was not initialized with a constant expression
         const size_t column_size = list_.begin()->size();

And yes, I know that the length of each column may be different, but I've stripped out some code for simplicity. Actually, I am coding a mathematical matrix data structure for C++. And I also know that the two-dimensional array can be treated as one-dimensional and can be initialized easily using one-dimensional initializer_list.

How do I bypass this error? And why is this error present here?

Evg , I think that new T[n][m] is allocating a new two-dimensional array of order n*m — Artaza Sameen
– Artaza Sameen, Commented Sep 4, 2021 at 10:13
Take a look: en.cppreference.com/w/cpp/language/new#Explanation — Evg
– Evg, Commented Sep 4, 2021 at 10:17
Constructor TwoArray(initializer_list<initializer_list<int>> list_) is private. — pvc
– pvc, Commented Sep 4, 2021 at 14:34

IWonderWhatThisAPIDoes · Accepted Answer · 2021-09-04 10:19:51Z

1

You need to create one array of pointers first, then initialize them with arrays of int. Note that this may leak if any allocation throws an exception.

array = new int*[row_size];
for(std::size_t i = 0; i < row_size; ++i)
    array[i] = new int[column_size]{};

And then analogously delete each sub-array.

for(std::size_t i = 0; i < row_size; ++i)
    delete[] array[i];
delete[] array;

Although, unless you benefit from owning the memory directly, consider using std::vector<std::vector<int>> (or std::unique_ptr<std::unique_ptr<int[]>[]>) instead.

edited Sep 4, 2021 at 10:19

answered Sep 4, 2021 at 10:17

IWonderWhatThisAPIDoes

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1 Comment

Evg Over a year ago

Or even better, linear std::vector<int> with some index arithmetic.

Pepijn Kramer · Accepted Answer · 2021-09-04 12:59:54Z

I made this example so you can type:

array<int, 2, 3> my_array{ {{0,1}, {2,3}, {4,5}} };

Not fully tested yet but it will get you started. It also shows how you can achieve it without needing "new" or "delete".

#include <cassert>
#include <array>

template<typename type_t, size_t COLS, size_t ROWS>
struct array
{
    array(const type_t(&values)[ROWS][COLS])
    {
        for (auto c = 0; c < COLS; ++c)
        {
            for (auto r = 0; r < ROWS; ++r)
            {
                at(c, r) = values[r][c];
            }
        }
    }

    type_t& at(const std::size_t column, const std::size_t row)
    {
        assert(row < m_rows);           // row out of range
        assert(column < m_columns);     // column out of range

        auto index = (row * COLS) + column;
        return m_data[index];
    }

    type_t& operator()(const std::size_t column, const std::size_t row)
    {
        return at(column, row);
    }

private:
    const type_t m_rows{ ROWS };
    const type_t m_columns{ COLS };
    std::array<type_t, ROWS* COLS> m_data{};
};

int main()
{
    array<int, 2, 3> arr{ {{0,1}, {2,3}, {4,5}} };

    assert(arr(1, 1) == 3);
    assert(arr(1, 2) == 5);
}

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