How to correctly pass an 2D array pointer as argument?

Question

For the heads up: I am incredibly new with cpp. Being used to PHP and JS as work languages, I find pointers incredibly confusing.

So I have this Class called Macierz. The class is supposed to hold Matrixes of float variables and ought to have a constructor accepting an 2d array in order to print them into the field.

The field is declared like this float mx3[3][3];

And the constructor has such declaration: Macierz(float**); With the body using an additional function:

Macierz::Macierz(float** f) {
    length = 3;
    populateWith(f, length);
}

void Macierz::populateWith(float** val, int length) {
    for (int i = 0; i < length; i++)
        for (int j = 0; j < length; j++)
            mx3[i][j] = val[i][j];
}

In my main() function I want to declare the class with a created float array. I try to do it as so, but It just won't work the way God intended:

float y[3][3] = { {1.00f, 2.00f, 3.00f}, { 4.00f, 5.00f, 6.00f }, { 7.00f, 8.00f, 9.00f } };

Macierz m5(y);

This is the base template of what I want to do. I've tried making the y variable a double pointer, a regular pointer, passing via reference and it just won't kick.

What would be the most prober way to pass this variable?

Any help will be amazing, I am really a noob in this language.

An array of arrays is not equivalent to a pointer to pointer. See e.g. [this old answer of mine](https://stackoverflow.com/questions/18440205/casting-void-to-2d-array-of-int-c/18440456#18440456) for a simple drawing of the difference. — Some programmer dude, Apr 12 '18 at 07:50
And to help you I suggest you stat using the standard library and its [containers](http://en.cppreference.com/w/cpp/container). In your case, since the arrays is of a compile-time fixed size you could use [`std::array`](http://en.cppreference.com/w/cpp/container/array). Something like `std::array, 3>, 3> y = { ... }`. Then just pass references to the type as arguments and copy it as one would normally copy a variable (using assignment). — Some programmer dude, Apr 12 '18 at 07:53
@Someprogrammerdude For now the project has many elements in it using the float[][] type thing, therefore changing the base way it works would add alot more work into it. If I had the chance now, Id do it with Vectors probably :P But thanks for the help — aln447, Apr 12 '18 at 07:57

score 1 · Answer 1 · answered Apr 12 '18 at 08:00

1

You need to remember that arrays naturally decays to pointers to their first element. But that isn't recursive.

For y in your example, the decay is to a pointer to the first element of y, which is equal to &y[0]. That is a pointer to an array, and will have the type float(*)[3], which is the type you need for your arguments:

Macierz::Macierz(float (*f)[3]) { ... }

answered Apr 12 '18 at 08:00

Some programmer dude

400,186
35
402
621

Thanks. I have tho found another way by rewriting the table into a pointer to pointers one. Check my answer below. Turned out to be quite of a basic answer :) – aln447 Apr 12 '18 at 08:06

score 0 · Answer 2 · answered Apr 12 '18 at 08:07

0

Managed to fix this by copying the float into a pointer-to-pointer types

float y[3][3] = { {1.00f, 2.00f, 3.00f}, { 4.00f, 5.00f, 6.00f }, { 7.00f, 8.00f, 9.00f } };

    float** x = 0;
    x = new float*[3];

    for (int i = 0; i < 3; i++) {
        x[i] = new float[3];

        for (int j = 0; j < 3; j++) {
            x[i][j] = y[i][j];
        }
    }

    Macierz m5(x);

answered Apr 12 '18 at 08:07

aln447

981
2
15
44

2

Don't forget to `delete[]` the memory you `new[]`. Its also an extra step which of course will have an impact on performance. And you don't need the actual copying of the values, just do e.g. `x[i] = y[i]` in the outer loop. – Some programmer dude Apr 12 '18 at 08:21
Sure. Thank you! – aln447 Apr 12 '18 at 08:47

How to correctly pass an 2D array pointer as argument?

2 Answers2