Use of & in declaring pointers to array

user_19681418 0
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To make a pointer to a whole array we proceed like that:

    int arr[3] = {1,2,3};
    int (*p)[3] = &arr;

How come i get an incompatibility error when trying to do the same with a 2D array?

    int arr[3][3] = {{12,10,45}, {44,55,66}, {79,85,91}};
    int (*p)[3] = &arr;

The problem here is &.

I'm sure this is a simple question that might have already been answered but i don't find any answer to this specific issue concerning the use of &.

pointers c
3 Answers

You've got p in the second example as a pointer to a 1D array, not a pointer to a 2D array. For that, you need the following:

int (*p)[3][3] = &arr;
user_16871190

In C, "pointer" is a category of types. To get a particular type, you have to specify what type of object the pointer points to.

Accordingly, the unary & does not generically compute a pointer. It creates a pointer to the type of its operand.

Similarly, in C, "array" is a category of types. To get a particular type, you have to specify the element type, the number of dimensions, and the sizes of at least the last n-1 dimensions.

Thus, with

    int arr[3] = {1,2,3};
    int (*p)[3] = &arr;
  • arr is defined as an array of 3 int, therefore
  • &arr has type pointer to array of 3 int (spelled int (*)[3] as a C type name), and
  • p is declared with that type

so everything is consistent.

How come i get an incompatibility error when trying to do the same with a 2D array?

Because your "the same" is not analogous.

With

    int arr[3][3] = {{12,10,45}, {44,55,66}, {79,85,91}};
    int (*p)[3] = &arr;

, arr has type array of 3 array of 3 int, and &arr has type pointer to array of 3 array of three int (a.k.a. int (*)[3][3]). This is not the same type that p has. The appropriate declaration of p for this version of arr would be

    int (*p)[3][3] = &arr;
user_24022720

arr has type int [3][3], so &arr has type int (*)[3][3]. So you can use:

int arr[3][3] = {{12,10,45}, {44,55,66}, {79,85,91}};
int (*p)[3][3] = &arr;

You can then access p as (*p)[i][j].

But it would be more typical to use:

int arr[3][3] = {{12,10,45}, {44,55,66}, {79,85,91}};
int (*p)[3] = arr;

In this case, int [3][3] is compatible with an int (*)[3] pointer, and you can use p[i][j] when accessing it.

An analogous situation exists with one-dimensional arrays. If you have int x[3], then the usual pointer type to use is int *p, in which case you would do p = x and access it as p[i].

But if you really want the extra level of indirection, you could use:

int (*p)[3] = &x;

Then instead of p[i], you would now need to use (*p)[i] when accessing it.

user_54607190
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