Re: Lets put it another way

From:

Ian Collins <ian-news@hotmail.com>

Newsgroups:

comp.lang.c++

Date:

Thu, 26 May 2011 09:57:38 +1200

Message-ID:

<945caiFn9oU3@mid.individual.net>

On 05/26/11 09:39 AM, Paul wrote:

"Ian Collins"<ian-news@hotmail.com> wrote in message
news:945a35Fn9oU1@mid.individual.net...

On 05/25/11 10:13 PM, Paul wrote:

"Ian Collins"<ian-news@hotmail.com> wrote in message
news:942v9uFv4kU3@mid.individual.net...

On 05/25/11 08:11 AM, Paul wrote:

"Ian Collins"<ian-news@hotmail.com> wrote:

On 05/25/11 12:58 AM, Paul wrote:

You are not addressing the what stores the memory address.

Because there isn't an address to store. The typeid operator (like
sizeof) works with types, not values. In a simple case such as this
the
operand is not evaluated. Until you understand this you will be stuck
in
limbo.

I could have simplified my simplification further by writing

std::cout<< typeid(int[3]).name()<< std::endl;

See? No memory address just a type.

There is an address stored.

Look at the code :

int main(){
int (*pp)[3] = (int (*)[3])new int[3];
std::cout<<*pp;
}
If this does not output an address what the hell is it?

The value returned by new.

Which is an address value.

Correct, an int*.

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?

There isn't an address. Where's the address in typeid(int).name()?
Or
sizeof(int)?

The address is shown in my code above.

The address in your code id the value return by new. No amount of
casting
will change that.

The value returned from new is a pointer value that is an address.

Why do you think it is not an address?

I don't. You changed the context. You originally posted

std::cout<< typeid(*p2).name();

No I originally posted:
  pparr p1 = (pparr)new int[3];
  pparr p2= p1;
  delete[] p1;
  std::cout<< *p2<<std::endl;

You conveniently omitted the last line:

std::cout<< typeid(*p2).name();

The code I posted output a memory address.

That is one possible outcome of UB. It could also have output garbage,
or if you have been using one of my specialised x86 debug allocators,
aborted the process.

and asked "what is the object that stores the address and is interpreted
as an array type by the typeid expression?" and I explained that no object
is required for typeid because the type is known at compile time. Did you
understand that?

typeid does not require an object but when it is passed an object it gives
us the type of that object.

It isn't passed an object. It gives us the *static* type of the
expression it is applied to. The exception is when the expression is
polymorphic class type.

--
Ian Collins