Re: Problem with array objects

From:

Ian Collins <ian-news@hotmail.com>

Newsgroups:

comp.lang.c++

Date:

Sun, 24 Apr 2011 14:10:47 +1200

Message-ID:

<91hf57F6cmU6@mid.individual.net>

On 04/24/11 01:16 PM, Paul wrote:

"Ian Collins"<ian-news@hotmail.com> wrote in message
news:91h5dpF6cmU3@mid.individual.net...

On 04/24/11 11:04 AM, Paul wrote:

A pointer to an array of 4 ints, when dereferenced, returns an array
of 4 ints. Look at the output of running the following program.

    #include<iostream>
    #include<typeinfo>

    int main() {
      int a1i[4], (*pa1i)[4] =&a1i;

      std::cout<<"typeid( a1i)="<<typeid(a1i).name()<<std::endl;
      std::cout<<"typeid(*pa1i)="<<typeid(*pa1i).name()<<std::endl;
    }

You are dereferencing an array-type , not an array.
For example:

int arr[40] ={0};
    int* parr = arr;
    int (*pparr)[40] =&arr;

    std::cout<< "address of arr[0]:\t"<< &arr[0]<<std::endl;
    std::cout<< "value of *parr:\t\t"<< *parr<<std::endl;
    std::cout<< "value of *pparr\t\t"<< *pparr<<std::endl;

Dereferencing pparr does not access the array it accesses a temporary
pointer.
The pointer pparr does not reference the array, it references an
array-type
object.

Dereferencing pparr accesses an array in the same sense that using
the name arr accesses an array.

No it doesn't

What I wrote was:
No it doesn't
*arr =1; /*Accesses the array*/
*pparr=1; /*Cannot do this*/

There is no need to snip something so short. Obviously you are attempting
to
change the context of my post from that of a reasonable debate into a
blunt
non explanatory argument.

I was replying to text above, not the example, which had no relevance to
the text.

Saying "*pparr=1; /*Cannot do this*/" is no different form writing

struct X { int n; };

   X x;
   X* px =&x;
   *px = 1;

Saying "*pparr=1; /*Cannot do this*/". is completely different from writing
the above code.
pparr's type is not a pointer to an X. It's type is pointer to array of
ints.

Exactly.

try adding

    (*pparr)[0] = 42;
    std::cout<< "value of arr[0]\t\t"<< arr[0]<<std::endl;
    arr[10] = 1234;
    std::cout<< "value of (*pparr)[0]\t\t"<< (*pparr)[10]<<std::endl;

This code only confirms that *arr, is not the same as *pparr, in any
sense.
Which is exactly what I said.

Did anyone dispute that? Remember you disputed "Dereferencing pparr
accesses an array in the same sense that using the name arr accesses an
array".

dereferencing pparr does not access the array , it..*drumroll* dereferences
pparr.

Neither does arr. Remember the origin claim you disputed?
"Dereferencing pparr accesses an array in the same sense that using the
name arr accesses an array".

So how does "(*pparr)[0] = 42;" differ from "arr[0] = 42;"?

If you can't see that they differ in levels of indirection, then you do not
understand the code.

Quite. Remember the words "Dereferencing pparr'?

Does (*pparr) not dereference pparr?

As I said dereferencing pparr doesn't access the arrray it accesses an
array-type object.
As can be shown:
std::cout<< *p;
std::cout<< typeid(*p).name();

Try adding

std::cout<< (typeid(arr) == typeid(*pparr))<< std::endl;

This is just a pointless comment that in no way suggests the *pparr access
the array.
This simply shows that *pparr has the same type as arr. I have already
stated that *pparr yields an array-type object.

So dereferencing pparr allows you to access the array in the same manner
as arr. You have just disproved your own claim.

Well done.

--
Ian Collins