Question

yet another beginner-to-intermediate question. I’m trying to pass a 2-D array to a function in C++. I’m aware that the array can’t be sent directly to the function, so I first created a pointer (names edited but you’ll get the idea):

double input[a][b] = {{0, 0}, {0, 1}, {1, 0}, {1, 1}};
Class.calculate (*(input), a, b);

Then I try to use it in said function – but seemingly I’m unaware on how to dereference the pointer to be able to handle the 2-D array again. My code (more or less):

for (int x=0; x<=a; x++){
    for (int y=0; y<=b; y++){
        tmpInput[x][y]= (*input)[x][y];
    }
}

The compiler complains about an error, namely invalid types ‘double[int]’ for array subscript, but I still can’t figure out the problem. My best bet is that I didn’t dereference the 2-D array properly, but the other option is that C++ can’t dereference 2-D arrays directly, instead relying in converting the array to 1-D before sending it. Any ideas?

Answer 1

It’s hard to say for sure without seeing the declaration of calculate(), but I’d guess it’s an argument type mismatch. You can send your array as a double*, but not as a double** – it is an array of doubles (they’re laid out linearly in memory, even if the compiler lets you treat them as a multidimensional array). But it is not an array of pointers to doubles, which is what the double** type actually means: Dereference a double** and you get a double*.

If you know that the second subscript is constant (in your case, 2), you can define calculate() to expect a N*2 array of doubles:

void calculate(double array[][2], int a)
{
    //Whatever
}

Or, if both dimensions can change, just take a straight double* and to the pointer math yourself (you’ll probably have to cast your array to a double* when calling this, but it’ll work):

void calculate(double* array, int a, int b)
{
    //To get array[i][j]:
    array[i * b + j];
}

Hope this helps!