Doesn't a 2D array decay to pointer to pointer

Question

Up till now I was pretty much sure that

int arr[4][5];

Then arr will decay to pointer to pointer.

But this link proves me wrong.

I am not sure how did I get about arr being pointer to pointer but it seemed pretty obvious to me. Because arr[i] would be a pointer, and hence arr should be a pointer to pointer.

Am I missing out on something.

Answer 1

Yep you are missing out on a lot :)

To avoid another wall of text I'll link to an answer I wrote earlier today explaining multi-dimensional arrays.

With that in mind, arr is a 1-D array with 4 elements, each of which is an array of 5 ints. When used in an expression other than &arr or sizeof arr, this decays to &arr[0]. But what is &arr[0]? It is a pointer, and importantly, an rvalue.

Since &arr[0] is a pointer, it can't decay further. (Arrays decay, pointers don't). Furthermore, it's an rvalue. Even if it could decay into a pointer, where would that pointer point? You can't point at an rvalue. (What is &(x+y) ? )

Another way of looking at it is to remember that int arr[4][5]; is a contiguous bloc of 20 ints, grouped into 4 lots of 5 within the compiler's mind, but with no special marking in memory at runtime.

If there were "double decay" then what would the int ** point to? It must point to an int * by definition. But where in memory is that int * ? There are certainly not a bunch of pointers hanging around in memory just in case this situation occurs.