CODEWITHSUNDEEP
carrayspointers
pistacchio
pistacchio

Reputation: 58963

Array of pointers to multidimensional arrays

i have some bidimensional arrays like:

int shape1[3][5] =  {1,0,0,
             1,0,0,
             1,0,0,
             1,0,0,
             1,0,0};
int shape2[3][5] =  {0,0,0,
             0,0,0,
             0,1,1,
             1,1,0,
             0,1,0};

and so on.

How can i make an array of pointers to those?

I tried the following, but they don't work (warning: initialization from incompatible pointer type):

int *shapes[]=  {&shape1,&shape2};

int *shapes[]=  {shape1,shape2};

int **shapes[]= {&shape1,shape2};

Any help?

Upvotes: 4

Views: 7876

Answers (3)

dirkgently
dirkgently

Reputation: 111336

Updated Fixed type. Thanks j_radom_hacker for bringing this to my attention!

[EDIT: Actually the type here was not correct -- see Robert S. Barnes' answer for the correct type to use.]

Figure out the type of shape1 and shape2 first:

typedef int (*shape_array_t)[5];

Now use this:

shape_array_t sat[] = { shape1, shape2 };

Upvotes: 3

j_random_hacker
j_random_hacker

Reputation: 51326

First of all, the first array bound refers to the outermost array dimension, so you should probably declare shape1 as:

int shape1[5][3] =  {1,0,0,
                     1,0,0,
                     1,0,0,
                     1,0,0,
                     1,0,0};

and similarly for shape2.

[EDIT: I've changed the type of shapes below to correspond to Robert Barnes' answer -- we don't want the outermost subscript to be included in this type!]

The slightly strange-looking typename you need is:

int (*shapes[])[3] = { shape1, shape2 };

This allows the element at row 4, column 1 of shape2 to be addressed using

shapes[1][3][0]

Breakdown of subexpressions and their C types:

shapes            // has type "int (*x[2])[3]" (decays to "(**x)[3]")
shapes[1]         // has type "int (*x)[3]"
shapes[1][3]      // has type "int x[3]" (decays to "int *x")
shapes[1][3][0]   // has type "int x"

(Note that a dummy x has been included in the types above to make them clearer -- in fact this identifier is not part of the type.)

A rule of thumb for decoding C/C++ types is "starting from the variable name, read right when you can and left when you hit a closing parenthesis." So the decoded typename for shapes is:

An array of pointers to an array of 3 integers.

In general it's much nicer to use typedefs for these complicated types, as dirkgently suggests.

Upvotes: 3

Robert S. Barnes
Robert S. Barnes

Reputation: 40598

I believe I just verified what I wrote was correct. The following works as expected:

#include <stdio.h>

int main(int argc, char **argv) {

int shape1[5][3] =  {1,0,0,
                 1,0,0,
                 1,0,0,
                 1,0,0,
                 1,0,0};

int shape2[5][3] =  {0,0,0,
                 0,0,0,
                 0,1,1,
                 1,1,0,
                 0,1,0};

typedef int (*shapes_p)[3];
shapes_p shapes[2] = { shape1, shape2 };

shapes[0][1][0] = 5;
shapes[1][1][0] = 5;

printf("shape1[1][0] == %d\n", shape1[1][0]);
printf("shape2[1][0] == %d\n", shape2[1][0]);

}

The thing to remember is that the type of shape1 and shape2 is actually:

int *shape1[5];

What you have in memory is 3 adjacent arrays of 5 ints each. But the actual type is pointer to array of 5 ints. When you write:

shape1[1][2] = 1;

you're telling the compiler to index to the second array of int[5] and then access the 3rd element of that array. What the compiler actually does is pointer arithmetic on the underlying type pointed to, in this case int[5]. You could do the same with the following code:

int *p = shapes1[0];
p+7 = 1;  // same as shape1[1][2] = 1;

So if you want an array of pointers to int *[5] then you would do:

typedef int (*shapes_p)[5];
shapes_p shapes[2];

Upvotes: 6

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