Re: Problem with array objects
"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.
All together now - oh yes it does!
You seem to have nothing positive to add as an individual. All you attempt
to do is create a "more is better" argument.
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.
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();
*p = 50; /*Wrong. cannot access the array*/
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.
For example:
(i) int* p = *p;
(ii) int (&r)[40] = *p;
These are the only two object-types that can access the array, namely:
(i) A pointer to an array.
(ii) A reference to an array.