Pointers equivalent to Arrays

Question

I know that If i define an array like

int a [10];

I can use a pointer notation, to access it's address using a+<corresponding_item_in_array> and it's value using, *(a+<corresponding_item_in_array>) .

Now I wanted to reverse things, I used malloc to allocate a memory to a integer pointer, and tried to represent the pointer in subscript notation but it didn't work

int *output_array;
output_array = (int *) (malloc(2*2*2*sizeof(int))); //i.e, space for 3d array

output_array[0][0][1] = 25;  
// ^ produces error: subscripted value is neither array nor pointer

I may have used an pointer expression using Storage Mapping, but ain't the simpler method available? and Why?

You can use the answer at this post to extend to 3rd dimension as well: http://stackoverflow.com/questions/1052818/create-a-pointer-to-two-dimensional-array — Tuxdude, Feb 28 '13 at 01:55

score 3 · Accepted Answer · answered Feb 28 '13 at 02:00

3

The int* type is not an equivalent of the 3D array type; it is an equivalent of a 1D array type:

int *output_array;
output_array = (int *) (malloc(8*sizeof(int))); //i.e, space for array of 8 ints
output_array[5] = 25; // This will work

The problem with arrays of higher ranks is that in order to index into a 2D, 3D, etc. array the compiler must know the size of each dimension except the first one in order to calculate the offsets from the indexes correctly. To deal with 3D arrays, define a 2D element, like this:

typedef int element2d[2][2];

Now you can do this:

element2d *output_array;
output_array = (element2d*) (malloc(2*sizeof(element2d))); 
output_array[0][0][1] = 25; // This will work now

Demo on ideone.

answered Feb 28 '13 at 02:00

Sergey Kalinichenko

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Does tha `malloc`ed, element be on the top, i mean, output_array[malloced][2][2] ? – cipher Feb 28 '13 at 02:10
1

Incidentally, you can do it even without the typedef, but the syntax is way more awkward (the declaration would be `int (*outputArray)[2][2];` and the cast would be `(int (*)[2][2])`). – Matteo Italia Feb 28 '13 at 02:18
@cipher Correct, it's on the inside. So if you do `malloc(200*sizeof(element2d))` the highest index would be `outputArray[199][1][1]`. – Sergey Kalinichenko Feb 28 '13 at 02:34

autistic · Answer 2 · 2022-06-29T03:57:19.137

What's the type of output_array? int *.

What's the type of *(output_array+n) or output[n]? int.

Is subscript permitted on int? Both subscripts (eg. *(output_array+n) and output[n]) are pointer operations, and int is not a pointer. This explains the error you recieved.

You can declare a pointer to int[x][y] like this: int (*array)[x][y];

You can allocate storage that's a suitable alternative to a 3D array to array using: array = malloc(42 * x * y);. This would be the equivalent to int array[42][x][y];, except that arrays aren't modifiable lvalues, the alignof, sizeof and address-of operators work differently and the storage duration is different.

score 0 · Answer 3 · answered Feb 28 '13 at 02:01

Because compiler doesn't have any idea about the size of each dimension, so it's not able to find out where output_array[0][0][1] should be.

You could try this

typedef int (* array3d)[2][2];
array3d output_array;
output_array = (array3d) malloc(2 * 2 * 2 * sizeof(int));
output_array[0][0][1] = 25;
printf("%d\n", output_array[0][0][1]);

Pointers equivalent to Arrays

3 Answers3