Passing 2D Array of Pointers in C

Question

For my program I need to pass a 2D array of pointers to a function in a separate file. I've written a similarly-syntaxed file below:

#include <stdlib.h>
#include <stdio.h>

typedef struct {
  int state;
  int design;
} card_t;

card_t *cardSet[5][5];

void setFirst(card_t *cards[][]) { // <- Error: Array has incomplete element type
  cards[0][0]->state = 1;
}

int main() {
  setFirst(cardSet);  // <- Error: Type of formal parameter 1 is incomplete
  return 0;
}

When I change the code to all 1D arrays it compiles fine, but for a 2D array I get the errors shown above. What is the difference between the two cases?

Thanks! Cameron

MByD · Accepted Answer · 2011-04-08 23:47:36Z

1

if you pass an array to a function, you have to specify the size of the inner array, in your case, instead of void setFirst(card_t *cards[][]), you should specify void setFirst(card_t *cards[][5]).

Why do you need to specify it and not the size of the first dimension?

Since cards is an array of array of card_t pointers, if you want to get to cards[1][0], the compiler will need to know how much to add to the pointer cards - cards is declared: card_t *cards[5][4] it will need to add 4 * sizeof(*card_t) to get to cards[1][0], but if cards is declared: card_t *cards[5][5] it will need to add 5 * sizeof(*card_t) to get to cards[1][0].

edited Apr 8, 2011 at 23:47

answered Apr 8, 2011 at 23:35

MByD

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Comments

Ben Stott · Accepted Answer · 2011-04-09 06:41:06Z

1

As has been mentioned, to pass a 2d array to a function, you need to have every dimension but the first declared. However, you can also just pass the pointer, as follows. Note that you should always (unless the array dimension is completely fixed and the function that operates on the array only operates within the array's dimension) pass the length of each array, too.

void setFirst(card_t ***cards, size_t n, size_t m) {
  if (n > 0 && m > 0) {
    cards[0][0]->state = 1;
  }
}

Because referencing an array via code[0][0] is the same as *(*(code+0)+0*m), you can pass two pointers instead of array dimensions.

answered Apr 9, 2011 at 6:41

Ben Stott

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Comments

Mahesh · Accepted Answer · 2011-04-08 23:35:57Z

0

Only the first index is optional. You should definitely mention the second index, because a two dimensional array decays to a pointer to 1d array.

void setFirst(card_t *cards[][5]) {
                           // ^ Newly added

 // ..

}

Also make sure that the pointers are pointing to valid memory locations. Else dereferencing leads to segmentation faults. BTW, is there any reason to have a two dimensional array with pointers. I think you are just complicating the program.

answered Apr 8, 2011 at 23:35

Mahesh

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Comments

Jens Gustedt · Accepted Answer · 2011-04-09 06:31:43Z

0

To achieve something similar to what you are trying C99 has variable length arrays. These come particularly handy in function definitions:

void setFirst(size_t n, size_t m, card_t *cards[n][m]) {
...
}

Observe that you have to have the size parameters known at the moment of the array declaration, so you'd have to put them in front.

answered Apr 9, 2011 at 6:31

Jens Gustedt

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