How can I pass the pointer of a static 2D array to a structure/class?

Question

I have a problem when I try to pass the pointer of an array (which contains parameters needed by some functions in my program) to a structure, which then should be passed to those functions. GSL for example wants me to pass parameters in this way.

A little example program looks like this:

#include <iostream>

using namespace std;

struct myparams
{
    double  * a;
    double ** b;
};

int main()
{
    double c[10]   = {0,1,2,3,4,5,6,7,8,9};
    double d[4][3] = {{1,2,3},{4,5,6},{7,8,9},{10,11,12}};

    double** e = new double*[4];
    for (int i = 0; i < 4; i++) {
       e[i] = new double[3];
    }

    myparams params;

    // THIS WORKS:
    params.a = c;
    for (int i = 0; i < 10; i++) {
        cout << params.a[i] << endl;
    }

    // THIS DOESN'T WORK
    params.b = d;

    // THIS WORKS:
    params.b = e;

    delete[] e;
}

What is the problem with

 params.b = d

The Compiler complains with "cannot convert 'double[4][3]' to 'double**' in assignment" or something like that (translated from german).

`double[n][m]` cannot decay into `double **`, while `double[n]` can decay into `double *`. That's why the first assignment is valid, and the latter isn't. — Zeta, Feb 15 '13 at 14:08
Also see http://stackoverflow.com/a/13273979/560648 and http://stackoverflow.com/a/9557855/560648 and http://stackoverflow.com/a/7139014/560648 — Lightness Races in Orbit, Feb 15 '13 at 14:15

score 9 · Accepted Answer · edited May 23 '17 at 12:03

9

double d[4][3]; is an array of arrays. double b**; is a pointer to pointer. These types are not the same (the common "arrays are just pointers" you might have read on the internet is wrong).

Elements of d are of type double[3]. Arrays, when passed around, decay to pointers to their first element (see section 4.2. of C++ standard). d will decay to double(*)[3] (a pointer to array of 3 doubles).

Long story short, double(*)[3] is not convertible to double** and this is what compiler is telling you.

If you need to keep d as it is, you need to declare b as double (*b)[3];

For more in-depth explanation, refer to this SO question.

edited May 23 '17 at 12:03

Community

1
1

answered Feb 15 '13 at 14:16

jrok

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Thank you! This was almost the solution. It's not `double (*b)[4]`, but `double (*b)[3]`. – GriffinPeterson Feb 15 '13 at 14:20
I like how you accepted the incorrect answer over the correct one. That makes sense. – Lightness Races in Orbit Feb 15 '13 at 14:21
@LightnessRacesinOrbit If you point out what's incorrect, I'll fix it. Thank you. – jrok Feb 15 '13 at 14:22
@jrok: GriffinPeterson (the OP!) already did. Read his comment. – Lightness Races in Orbit Feb 15 '13 at 14:22
@LightnessRacesinOrbit Aye, typo. – jrok Feb 15 '13 at 14:23

Some programmer dude · Answer 2 · 2013-02-15T14:14:20.520

7

It's because a pointer to a pointer (or an array of pointers as it can be used as) is not the same as an array of arrays. The layout in memory is incompatible.

For example, lets say we have these two declarations:

char aa[2][2] = { { 'a', 'b' }, { 'c', 'd' } };
char **pp;

pp = new char*[2];
pp[0] = new char[2];
pp[1] = new char[2];

The array aa looks like this in memory:

+----------+----------+----------+----------+
| aa[0][0] | aa[0][1] | aa[1][0] | aa[1][1] |
+----------+----------+----------+----------+

The "array" pp meanwhile looks like this:

+------+------+
| a[0] | a[1] |
+------+------+
   |      |
   |      v
   |      +---------+---------+
   |      | a[0][0] | a[0][1] |
   |      +---------+---------+
   V
   +---------+---------+
   | a[0][0] | a[0][1] |
   +---------+---------+

The "array" pp contains pointers to another "array"

edited Feb 15 '13 at 14:14

answered Feb 15 '13 at 14:08

Some programmer dude

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Thank you for the quick response! But which member do I need in my structure `myparams` to store the pointer to the array `d`? – GriffinPeterson Feb 15 '13 at 14:14
Okay I think that's better than [my diagram](http://stackoverflow.com/a/7139014/560648). Dammit. – Lightness Races in Orbit Feb 15 '13 at 14:16
@LightnessRacesinOrbit Well, I think yours look more "professional" than my crude ASCII art :) – Some programmer dude Feb 15 '13 at 14:17
@JoachimPileborg: I did it in `pbrush`! – Lightness Races in Orbit Feb 15 '13 at 14:17
@GriffinPeterson: It's unwieldy, though. You're much better off with a 1D array representation. – Lightness Races in Orbit Feb 15 '13 at 14:29
@LightnessRacesinOrbit: Why unwieldy? In terms of speed? I already used a 1D array representation and arranged my indexing, but now I just wanted to know how to implement it in 2D. – GriffinPeterson Feb 15 '13 at 14:34
@GriffinPeterson: The type is unwieldy. Ownership semantics are unwieldy. It's just a mess. – Lightness Races in Orbit Feb 15 '13 at 14:37
I wish I could understand this – GriffinPeterson Feb 15 '13 at 14:44
@GriffinPeterson: http://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list – Lightness Races in Orbit Feb 15 '13 at 14:46
@GriffinPeterson In C++, there's really no need to use raw arrays anymore. [`std::vector`](http://en.cppreference.com/w/cpp/container/vector), or if a fixed size "array" is wanted [`std::array`](http://en.cppreference.com/w/cpp/container/array) is the way to go. Now you can even [initialize them like normal arrays](http://en.wikipedia.org/wiki/C++11#Initializer_lists). – Some programmer dude Feb 15 '13 at 15:51
Let `T b[3][5]; T *a = &b[0][0];` Why doesn't `*(a + j*5 + i)` resolve to `b[j][i]`? At least, that's what I'm getting from the diagrams but not actually trying it. – Noein May 04 '15 at 18:51

How can I pass the pointer of a static 2D array to a structure/class?

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