Re: Lets put it another way

From:
"Paul" <pchristor@yahoo.co.uk>
Newsgroups:
comp.lang.c++
Date:
Tue, 24 May 2011 13:58:39 +0100
Message-ID:
<aYNCp.9364$2E6.4304@newsfe18.ams2>
"Ian Collins" <ian-news@hotmail.com> wrote in message
news:941ffpFt2pU22@mid.individual.net...

On 05/24/11 09:20 PM, Paul wrote:

"Ian Collins"<ian-news@hotmail.com> wrote in message
news:940g82Ft2pU20@mid.individual.net...

On 05/24/11 01:14 PM, Paul wrote:

Ok in connection to my previous post, lets remove the null pointer
issue
and
consider this:

#include<iostream>

typedef int (*pparr)[3];

int main(){
   pparr p1 = (pparr)new int[3];
   pparr p2= p1;
   delete[] p1;
   std::cout<< *p2<<std::endl;


UB yet again.

   std::cout<< typeid(*p2).name();
}

In the last 2 lines does an array type object exist, even though there
is
no
array object?


No.

If not what is the object that stores the address and is interpreted as
an
array type by the typeid expression?


The typeid and sizeof operators do not evaluate their expression (unless
in case of typeid the type is polymorphic). So no object is required.


You do not seem to understand the question.


You do not appear to understand the answer.

I said ..if not what is the object that stores the address?
So what is this object? Do you know?


There isn't an object. The compiler knows the type of *p2. You may as
well have written

#include <iostream>
#include <typeinfo>

typedef int (*pparr)[3];

int main(){
  pparr p2;
  std::cout << typeid(*p2).name() << std::endl;
}


You are not addressing the what stores the memory address.
A memory address value is stored so there must be an object to store this
value, you say there is no object so what is it that stores this value?

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