Re: Problem with array objects
On 04/24/11 11:04 AM, Paul wrote:
"Ian Collins"<ian-news@hotmail.com> wrote in message
news:91guh2F6cmU1@mid.individual.net...
On 04/24/11 07:31 AM, Paul wrote:
A pointer to an array of 4 ints, when dereferenced, returns an array
of 4 ints. Look at the output of running the following program.
#include<iostream>
#include<typeinfo>
int main() {
int a1i[4], (*pa1i)[4] =&a1i;
std::cout<<"typeid( a1i)="<<typeid(a1i).name()<<std::endl;
std::cout<<"typeid(*pa1i)="<<typeid(*pa1i).name()<<std::endl;
}
You are dereferencing an array-type , not an array.
For example:
int arr[40] ={0};
int* parr = arr;
int (*pparr)[40] =&arr;
std::cout<< "address of arr[0]:\t"<< &arr[0]<<std::endl;
std::cout<< "value of *parr:\t\t"<< *parr<<std::endl;
std::cout<< "value of *pparr\t\t"<< *pparr<<std::endl;
Dereferencing pparr does not access the array it accesses a temporary
pointer.
The pointer pparr does not reference the array, it references an
array-type
object.
Dereferencing pparr accesses an array in the same sense that using
the name arr accesses an array.
No it doesn't
What I wrote was:
No it doesn't
*arr =1; /*Accesses the array*/
*pparr=1; /*Cannot do this*/
There is no need to snip something so short. Obviously you are attempting to
change the context of my post from that of a reasonable debate into a blunt
non explanatory argument.
I was replying to text above, not the example, which had no relevance to
the text.
Saying "*pparr=1; /*Cannot do this*/" is no different form writing
struct X { int n; };
X x;
X* px = &x;
*px = 1;
try adding
(*pparr)[0] = 42;
std::cout<< "value of arr[0]\t\t"<< arr[0]<<std::endl;
arr[10] = 1234;
std::cout<< "value of (*pparr)[0]\t\t"<< (*pparr)[10]<<std::endl;
This code only confirms that *arr, is not the same as *pparr, in any sense.
Which is exactly what I said.
Did anyone dispute that? Remember you disputed "Dereferencing pparr
accesses an array in the same sense that using the name arr accesses an
array".
So how does "(*pparr)[0] = 42;" differ from "arr[0] = 42;"?
Does (*pparr) not dereference pparr?
As I said dereferencing pparr doesn't access the arrray it accesses an
array-type object.
As can be shown:
std::cout<< *p;
std::cout<< typeid(*p).name();
Try adding
std::cout << (typeid(arr) == typeid(*pparr)) << std::endl;
*p = 50; /*Wrong. cannot access the array*/
Obviously (assuming p above is pparr). You have exactly the same
situation as the struct example above. But if you dereference the
pointer, you can use the result as if it were the original array, as my
additions and running the above showed you.
There are two ways this array-type object can be used to access the array
(i) Convert to a pointer.
(ii) Convert to a reference.
I think the examples above disprove that.
--
Ian Collins