Re: exercise on reinterpret_cast

From:
=?Utf-8?B?Rmls?= <Fil@discussions.microsoft.com>
Newsgroups:
microsoft.public.vc.language
Date:
Tue, 24 Jun 2008 14:10:01 -0700
Message-ID:
<AA25D0AE-A83E-4473-BFA4-A8091D7BF866@microsoft.com>
"Igor Tandetnik" wrote:

Fil <Fil@discussions.microsoft.com> wrote:

So I write p=iA.
But it doesn't work.

I thought I was doing something as natural as:
int a;
int * p;
p=&a;

Instead it works if I write p=&iA which makes no sense


Why? In the example above, you had to write p=&a, not p=a, and
apparently didn't find it the least bit surprising. So why do you have a
problem with p=&iA ?


Because iA should already contain the address of the first element of the
Array (like a fixed pointer).
Ex:
------------------------------------------------------------------
#include <iostream>
int main()
{
    int a[5];
    std::cout << a << " " << &a[0] << std::endl;
}
------------------------------------------------------------------
So if I am writing &iA it sounds to me like address of address of ...
This was confusing to me.

While I am talking I also tried to display for the first time &a and it
actually gives the same address than the 2 others. Waou.
 

iA is an address and I never asked to have it stored anywhere


I don't understand this at all. iA is an array. There's storage
allocated for this array. This storage begins at some address. &iA takes
this address.

(I mean
that an address of a pointer makes sense if the pointer has been
created.


First, iA is not a pointer, it's an array. Second, what you mean by
"created" here?

What also works is :
p=reinterpret_cast<double(*)[10]>(iA);
Is this the right way to do?


No, p=&iA is the right way to go (though declaring a pointer to an
array, as in double (*p)[10], is an unusual thing to want to do in the
first place).

static_cast doesn't work here: the compiler says:
cannot convert from 'double [10]' to 'double (*)[10]'
If I write ...
p=static_cast<double(*)[10]>(&iA[0]);
... I got : cannot convert from 'double *' to 'double (*)[10]'


That's pretty self-explanatory, isn't it?

This difference let me think that iA is more than &iA[0].


It is. Write a program that prints sizeof(iA) and sizeof(&iA[0]).

I maybe need to read something more thourough on this.
Can you please shed some light on the true nature of Arrays


http://www.transcendentaxis.com/dthompson/blog/archives/9

--
With best wishes,
    Igor Tandetnik

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necessarily a good idea. It is hard to be sure where they are going to
land, and it could be dangerous sitting under them as they fly
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