C: Pointer to 2-dimensional array of pointers

Question

I am trying to solve the following issue but could not succeed yet:

I have a two-diwmensional array of pointers:

int* a[16][128];

Now I want to make a pointer to this array in that way that I can use pointer arithmetic on it. Thus, something like this:

ptr = a;
if( ptr[6][4] == NULL )
  ptr[6][4] = another_ptr_to_int;

I tried already some variations but it either fails then on the first line or on the if condition.

So, how can it be solved? I would like to avoid template classes etc. Code is for a time critical part of an embedded application, and memory is very limited. Thus, I would like ptr to be only sizeof(int*) bytes long.

have you initialized all of `a` s elements? I.E a `int* a[16][5] = {NULL};` otherwise your computer may currently be [spray painting your cat purple](http://en.cppreference.com/w/cpp/language/ub). — George, Jan 25 '17 at 17:34
`auto ptr = a;` should do the job, so your `ptr` will have the correct type. — mch, Jan 25 '17 at 17:36
C and C++ are different languages. Choose **one**. That your compiler does not support `auto` as a type specifier suggests that C is the one you want to choose. — John Bollinger, Jan 25 '17 at 17:44

score 2 · Answer 1 · answered Jan 25 '17 at 17:45

2

A pointer to the first element of the array (which is what you want) could be declared as

int* (*ptr)[128];

A pointer to the array itself would be

int* (*ptr)[16][128];

and is not what you're looking for.

answered Jan 25 '17 at 17:45

molbdnilo

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Ok. But what is the size of ptr in memory then? sizeof(int*) or 128*sizeof(int*)? – dspverden Jan 25 '17 at 18:24
@dspverden You can check easily by printing `sizeof(ptr)`. (It's 99.99% certain that it will be the same as `sizeof(int*)`.) – molbdnilo Jan 25 '17 at 19:11
@dspverden It's extremely unlikely that an embedded device has pointers of varying size. (Only very esoteric hardware does it.) Unless your documentation states otherwise, it's safe to assume that `sizeof(T*) == sizeof(U*)` for all types `T` and `U`. – molbdnilo Jan 25 '17 at 21:25

score 1 · Answer 2 · answered Jan 25 '17 at 17:46

1

Thing you seem to want:

int* (*ptr)[128] = a;

Actual pointer to the array:

int* (*ptr)[16][128] = &a;

answered Jan 25 '17 at 17:46

yuri kilochek

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score 0 · Answer 3 · edited May 23 '17 at 12:16

To start with array pointer basics for a 1D array, [tutorialspoint][1] has a very easy to ready description. From their example:

    #include <stdio.h>

int main () {

   /* an array with 5 elements */
   double balance[5] = {1000.0, 2.0, 3.4, 17.0, 50.0};
   double *p;
   int i;

   p = balance;                                 //Here the pointer is assign to the start of the array

   /* output each array element's value */
   printf( "Array values using pointer\n");

   for ( i = 0; i < 5; i++ ) {
      printf("*(p + %d) : %f\n",  i, *(p + i) );
   }

   printf( "Array values using balance as address\n");

   for ( i = 0; i < 5; i++ ) {
      printf("*(balance + %d) : %f\n",  i, *(balance + i) );    // Note the post increment
   }

   return 0;
}

There are a couple of relavent Stack overflow answers that describe 2D arrays: How to use pointer expressions to access elements of a two-dimensional array in C?

Pointer-to-pointer dynamic two-dimensional array

how to assign two dimensional array to **pointer ?

Representing a two-dimensional array assignment as a pointer math?

  [1]: https://www.tutorialspoint.com/cprogramming/c_pointer_to_an_array.htm

C: Pointer to 2-dimensional array of pointers

3 Answers3