Discussion on: #001 DS&A - Operators and Arrays

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address of A[i] = i * sizeof(element) + base_address

An address is a pointer value, so when adding to it, the size is already factored in.

Which means that &A[i] is equivalent to A + i.

The only way that your formula works if it you're working with char *, in which case you'll get the wrong kind of pointer back at the end.

Beyond that, C doesn't have a flat memory model, so you can't just pretend that pointers are integers.

Two Dimensional Arrays

C does not support two dimensional arrays -- but it does support arrays of arrays.

So there's no need to transform it to a one dimensional array -- it already is one dimensional.

let's take this example of an 2D array with 3 rows and 4 columns

// An array of 3 int[4]s.
int array[3][4];

Indexing an array of arrays is simple pointer arithmetic.

Just as array[i] is (array + i), so is array[i][j] is *((array + i) + j).

It's a bit unclear to me why you're going out of your way to make this slower and more complicated. :)

Omar • Aug 23 '20 • Edited

Hi thanks for your comment ,
When you put A+i the compiler will convert it to A + (i*sizeof(element)) ,
for multi dimensional arrays yes they exist in C you can look at page 99 , section 5.7 of the original Ainsi C written by the Author of C.
But they are not the only way to do it, if you open in page 101 there is a section for Multi arrays vs pointers.
Tomorrow I will write about pointers :)
I am assuming that we don't know what is a pointer yet.

pentacular • Aug 23 '20

What he refers to as a multidimensional array is a one dimensional array of arrays.

Provide an example of something that you think is not a one dimensional array in C, and I will happily demonstrate to you why it is a one dimensional array. :)

If you don't know what a pointer is, you cannot meaningfully talk about arrays (since pointers are how you index arrays) or addresses (which are pointer values).

Omar • Aug 23 '20

Can you give me an academic resource to read it?
Also check this one lysator.liu.se/c/c-faq/c-2.html

pentacular • Aug 23 '20

Sure: open-std.org/jtc1/sc22/wg14/www/do... 6.5.2.1 Array subscripting

Lysator is generally confused about arrays -- consider the mess it makes of 2.14 which is a pointer into an array of int *, rather than being multidimensional.

The simplest proof is this.

int array[3][4];
int (*p)[4] = &array[0];
sizeof array[0] == sizeof (int[4]);

Therefore array is an array of length 3 with an element type of int[4].

Note the single dimension used to access it here.

Of course, having accessed array[i], you can now access that one dimensional array via array[i][j], and so on.

Probably the easiest way to think about it is that C provides arrays of arrays, which is one way to implement multi-dimensional arrays, without any of the arrays actually being multi-dimensional.

Omar • Aug 23 '20

it's all semantics there is no such thing as truly multidimensional arrays as all arrays are one dimensional
you can use a 2 or n dimensional accessor but that's about it.
there is no need to layer a simple thing to look complex.

pentacular • Aug 24 '20

There are such things are truly multi-dimensional arrays, just not in C.

For example, in pascal

type
  StatusType = (X, O, Blank);
  BoardType = array[1..3,1..3] of StatusType;
var
  Board : BoardType;

for count1 := 1 to 3 do
  for count2 := 1 to 3 do
    Board[count1, count2] := Blank;

Or in Common Lisp

(let ((a (make-array '(4 3))))
  (aref a i j))

This is not a criticism of C -- deciding to use arrays of arrays rather multi-dimensional arrays is a perfectly reasonable design choice.

It's just a design choice that should be understood clearly. :)