Once again Ian Collins is incorrect

[Paul]

In a recent discussion about a pointer to an array in the form of, int
(*pparr)[N]. The object that is a result of dereferencing such a pointer is
an array-type object. In this context the following was said:

I think the examples above disprove that.

What examples? And what do you suggest they disprove about the above?
If you disagree with what I said here , can you show me how to access an
array without using a pointer or a reference, or converting to one?
If you cannot provide an example of code to back up your statement is
nothing more than meaningless drivel.
</quote>

Ian Collins has disputed what I said about accessing the array with the
statement "I think the examples above disprove that."

He has since suggested that (*pparr)[N], is evidence that supports his
claim.
Can someone please inform Mr Ian Collins that the expression (*pparr)[4]
first dereferences pparr and then coverts the result into a pointer of type
int*, which is then dereferenced to access the array.

He does not seem to understand even the most basic C++ terminology and he
seems a bit stupid. It would greatly appreciated if someone could explain to
him how things work, because TBH I've had enough of his stupidity.

 

[Ian Collins]

In a recent discussion about a pointer to an array in the form of, int
(*pparr)[N]. The object that is a result of dereferencing such a pointer is
an array-type object.

As usual, Paul starts a smokescreen thread after loosing an argument. He
started by claiming the statement

"Dereferencing pparr accesses an array in the same sense that using the
name arr accesses an array".

was false and ended up contradicting him self nicely:
applies to the other.
the same way as arr

I'll leave it at that.

Sorry to impose yet another Paul thread on the group.

 

[Paul]

In a recent discussion about a pointer to an array in the form of, int
(*pparr)[N]. The object that is a result of dereferencing such a pointer
is
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.
With:
int arr[22];
int (*pparr)[22] = &arr;
Dereferencing those two identifiers does not access the array in the same
sense. There is clearly two different levels of indirection, as can be
shown:
*arr;
**pparr;

Are you completely stupid , or just partially stupid? You obviously have no
clue what you are talking about.

*plonk*

 

[Peter Remmers]

Am 25.04.2011 11:54, schrieb Paul:

In a recent discussion about a pointer to an array in the form of, int
(*pparr)[N]. The object that is a result of dereferencing such a pointer
is
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.
With:
int arr[22];
int (*pparr)[22] =&arr;
Dereferencing those two identifiers

The above statement only talks about dereferencing one of them. I've
already tried to tell you this in the other thread. The above version of
the statement actually has been reworded to more explicitly express what
it means.

Don't be so superficial. Read more closely. Understand a statement
before you dispute it and make a fool of yourself.

does not access the array in the same
sense. There is clearly two different levels of indirection, as can be
shown:
*arr;
**pparr;

Are you completely stupid , or just partially stupid? You obviously have no
clue what you are talking about.

*plonk*

You obviously have no clue that you are talking about.


Peter

 

[SG]

In a recent discussion about a pointer to an array in the form of, int
(*pparr)[N]. The object that is a result of dereferencing such a pointer
is
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.
With:
int arr[22];
int (*pparr)[22] = &arr;
Dereferencing those two identifiers does not access the array in the same
sense.

Nobody said it would. You gotta pay more attention. You seem to be
misunderstanding people constantly. Make sure that you understand what
Ian meant by

"using the name arr"

Hint: It does not imply putting a star in front of the name.

SG

 

[Paul]

Am 25.04.2011 11:54, schrieb Paul:


I think the examples above disprove that.

What examples? And what do you suggest they disprove about the above?
If you disagree with what I said here , can you show me how to access an
array without using a pointer or a reference, or converting to one?
If you cannot provide an example of code to back up your statement is
nothing more than meaningless drivel.
</quote>


Come on then idiot , lets see the code.
:

 

[Paul]

I think the examples above disprove that.

What examples? And what do you suggest they disprove about the above?
If you disagree with what I said here , can you show me how to access an
array without using a pointer or a reference, or converting to one?
If you cannot provide an example of code to back up your statement is
nothing more than meaningless drivel.
</quote>


Come on then idiot , lets see the code.
: