Re: Lets put it another way
"Joshua Maurice" <joshuamaurice@gmail.com> wrote in message
news:70ecc085-405f-41c0-b191-c29ea6f3f971@18g2000prd.googlegroups.com...
On May 24, 3:56 pm, "Paul" <pchris...@yahoo.co.uk> wrote:
"Joshua Maurice" <joshuamaur...@gmail.com> wrote in message
news:b6334244-434f-4c62-954b-c9b197253f8b@p13g2000yqh.googlegroups.com...
On May 24, 6:54 am, "Paul" <pchris...@yahoo.co.uk> wrote:
*p2 does not attempt to access the deallocated memory.
--Doesn't matter. It's UB. Attempting to read the pointer, not the
--deallocated memory, is UB. At least it is according to the C spec, and
--I expect that the C++ standard will follow suit if it hasn't already
This pointer obviously points to something that has a value that is
displayed on the screen as a memory address. It certainly doesn't point to
the integer objects that were allocated by new, so what does it point to?
How can you say it is UB if you do not first establish what it is pointing
to?
--Take the following program:
-- #include <stdlib.h>
-- #include <stdio.h>
-- int main()
-- { char* a = (char*)malloc(1);
-- char* b = a;
-- free(a);
-- printf("%p\n", b);
-- }
--This program has UB in C, and as I said will likely have UB in C++, or
--it already is UB in C++. See the defect report for more details:
--http://www.open-std.org/jtc1/sc22/wg14/www/docs/dr_260.htm
This is not the same as what I did.
--And I still don't know what you mean by array type object vs array
--object. Until you explain that technical distinction to me, you're
--literally just spouting effective nonsense, as you just invented those
--terms and no one else knows what you're saying.
Well consider the following:
int arr[3]={0};
std::cout<< arr;
The above does not output an array object, it outputs a memory address.
To output the array object you need to loop and dereference:
for (int i=0; i<3; ++i){
std::cout<< arr[i] << std::endl;
}
The array type object is a region of memory that contains a value that is
some memory address.
The array object is a region of memory than contains three integer
objects,
each with the value 0.
--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.
<snip>