Set a pointer to a static 2D array

Question

How can I set a pointer in a class to an external static data structure?

struct Str {
    double **matr;   // which type should matr be?
    int nx, ny;

    template<size_t rows, size_t cols>
    void Init(double(&m)[rows][cols], int sx, int sy) {
        matr = m;     // <-- error
        nx = sx; ny = sy;
    }
};
...
static double M[3][5] = { { 0.0, 1.0, 2.0, 3.0, 4.0 },
                          { 0.1, 1.1, 2.1, 3.1, 4.1 },
                          { 0.2, 1.2, 2.2, 3.2, 4.2 } };
Str s;
s.Init(M, 3, 5);

With this code, I get the following compile time error message (Visual C++ 2008/2012):

1> error C2440: '=' : cannot convert from 'double [3][5]' to 'double **'
1> Types pointed to are unrelated; conversion requires reinterpret_cast, C-style cast or function-style cast
1> see reference to function template instantiation 'void S::Init4<3,5>(double (&)[3][5],int,int)' being compiled

Your over-use of M is likely to get confusing - you have M the static array of doubles and M the template parameter for the void Str::Init() function. Which one are you intending double (*m)[M] to refer to? I think you want your mattr to be simply double *mattr, or maybe the uglier double (*mattr)[5]... — twalberg
– twalberg, Commented May 23, 2014 at 15:10
what does double(*m)[S] mean? Is it an array of function pointers? — Cool_Coder
– Cool_Coder, Commented May 23, 2014 at 15:16

ValarDohaeris-fuck-OpenAI · Accepted Answer · 2014-05-23 15:30:25Z

1

The problem is that a 2D array of double is not an array of pointers, it's just a single pointer to the first element of the 2D array that is represented by several contiguous rows of doubles in memory.

Since your struct has the fields nx/ny, you could just convert the array to a simple pointer and then use nx/ny to access it, i.e.:

struct Str {
    double *matr;
    int nx, ny;

    void Init(double* m, int sx, int sy) {
        matr = m;
        nx = sx; ny = sy;
    }
};

static double M[3][5] = { { 0.0, 1.0, 2.0, 3.0, 4.0 },
                          { 0.1, 1.1, 2.1, 3.1, 4.1 },
                          { 0.2, 1.2, 2.2, 3.2, 4.2 } };

int main() {
    Str s;
    s.Init(M[0], 3, 5);
    return 0;
}

You would then have to use nx/ny to access the array, e.g. here is a function that could be added to struct Str that prints the array:

#include <iostream>

void print() {
    for (int i = 0; i < nx; i++) {
        for (int j = 0; j < ny; j++) {
            std::cout << matr[i*ny+j] << " ";
        }
        std::cout << std::endl;
    }
}

Another (arguably nicer) solution would be to add template parameters to struct Str that replace nx/ny, then the matr member could have a type that contains the dimensions.

answered May 23, 2014 at 15:30

ValarDohaeris-fuck-OpenAI

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1 Comment

Pietro Over a year ago

If I use template parameters in the structure, I must specify the dimensions when I define the object. Can I use templates on the member function to have the option to resize the object later? Reference: user2079303's answer.

eerorika · Accepted Answer · 2014-05-23 15:34:26Z

1

So, you want to have a pointer to a 2D array. Str must be a template, because the type of it's member matr depends on the dimensions of that array.

template<int rows, int cols>
struct Str {
    double (*matr)[rows][cols];

    void Init(double(&m)[rows][cols]) {
        matr = &m;
    }
};

Str<3, 5> s;
s.Init(M);

answered May 23, 2014 at 15:34

eerorika

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