I have a function:
void foo(double[][4]);
which takes a 2d array with 2nd dimension equal to 4. How do I allocate a 2d array so that I can pass it to the function? If I do this:
double * arr[4];
arr = new double[n][4];
where n is not known to the compiler. I cannot get it to compile. If I use a generic 2d dynamic array, the function foo will not take it.
As asked, it is probably best to use a typedef
typedef double four[4];
four *arr; // equivalently double (*arr)[4];
arr = new four[n];
Without the typedef you get to be more cryptic
double (*arr)[4];
arr = new double [n][4];
You should really consider using standard containers (std::vector, etc) or containers of containers though.
typedef double v4[4];
v4* arr = new v4[n];
Consider switching to arrays and vectors though.
I know it may not be what OP has intended to do, but it may help others that need a similar answer.
You are trying to make a dynamic array of statically success array. The STL got your solution: std::vector and std::array
With these containers, things are easy easier:
std::vector<std::array<int, 4>> foo;
// Allocate memory
foo.reserve(8);
// Or instead of 8, you can use some runtime value
foo.reserve(someSize);
// Or did not allocated 8 + someSize, but ensured
// that vector has allocated at least someSize
// Add entries
foo.push_back({1, 2, 3, 4});
// Looping
for (auto&& arr : foo) {
arr[3] = 3;
}
// Access elements
foo[5][2] = 2;
Alternatively to creating a new type and occupying a symbol, you can create a pointer to pointer, and do it like that:
double **arr = new double*[j];
for (int i = 0; i < j; ++i)
{
arr[i] = new double[4];
}
whereas j is the int variable that holds the dynamic value.
I've written a simple code that shows it working, check it out here.
