Re: Lets put it another way

--I suggest you take a course in compilers, or read a book, like one of
--the Dragon Books, Red or Green. You have gross misunderstandings of
--both the C++ object model, and how compilers actually generate code.
--There is no such thing as an "array-type object", as you've defined
--it. In these examples, there are only array objects and pointer
--objects - including pointers to arrays and pointers to individual
--elements, not "array-type objects".

If there is no such thing as an array type object then what is the object
that is implicitly converted to a pointer, and stores a memory address?

--Consider:
-- void foo()
-- { int x[3];
-- int* y = x;
-- cout << y;
-- }

--"x" is a piece of text. It is a preprocessor token. It is an
--identifier for an auto (stack) variable. It names an object. It names
--an (array) object of type "int [3]". That array object has three sub-
--objects of type "int".

std::cout<< x;
//outputs a memory address.
std::cout<< typeid(x).name();
//outputs that x is an array type object.

"x" is an object that stores a memory address.
"x" is an array type object.

When the array to pointer conversion takes place what is converted to a
pointer?
It is not one of the three integer objects that is converted to a pointer,
it is the array type object.

--"y" is a piece of text. It is a preprocessor token. It is an
--identifier for an auto (stack) variable. It names an object. It names
--a (pointer) object of type "int*". It holds the address of the first
--element of the auto (stack) array x.

y is a pointer.

This function "uses" the following objects:
- the array object x, and its 3 unnamed int sub-objects,
- the pointer object y,
- the object cout.
-There is no "array-type object".

Thats your opinion but you do not acknoweldge the object that stores the
address of the array, the object that is converted to a pointer on
assignemt
to y.

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