Re: Lets put it another way

From:

"Paul" <pchristor@yahoo.co.uk>

Newsgroups:

comp.lang.c++

Date:

Wed, 25 May 2011 23:09:16 +0100

Message-ID:

<q6fDp.33063$IL2.3293@newsfe04.ams2>

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

No I didn't conveniently onit anything. I was proving the fact that the code
I originally posted output an address.

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.

So you say that *p2 is not an object? I can accept that I don't know what
the rules are which the reason for my OP.

But in argument to what you say, consider this:
pparr p1 = (pparr)new int[3];
pparr p2= p1;
std::cout<< *p2<<std::endl;
std::cout<< typeid(*p2).name();

Is *p2 an object now?