On 23/03/2011 17:24, Paul wrote:
On 23/03/2011 13:29, Paul wrote:
Odd multiples of > denotes Leigh bollocks..
Please enjoy:
You fail yet again. I never said "pa" was not a pointer
to an
array;
of course it is a pointer to an array; it is a pointer to
a 1d
array
of type int[42].
No its a pointer to a 2D array of objects of type int[42];
It's a pointer to a 2d array
According to you any dynamic array is, not unlike you, 1
dimension
short
of a picnic.
int (*pa)[42] = new int[42][42];
"pa" is a pointer to a 1D array of type int[42]; this is an
indisputable fact. The fact that what it points to is the
initial
element of a 2D array of type int[42][42] is irrelevant.
No its a pointer to a 2D array of objects of type int[42];
It's a pointer to....*drumroll* ..... what it points-to .
You are a confused idiot. Read the standards.
Follow your own advice (see below).
The standard states : "If the * operator, either explicitly or
implicitly as
a result of subscripting, is applied to this pointer, the
result is
the
pointedto (n - 1 )dimensional array"
............
(*pa) produces the **pointedto** (n-1) dimensional array. That
is....
if pa
is the **pointedto** 2dim array(which it is) then
derefenrencing it
should
result in a 1dim array.
std::cout<< typeid(*pa).name(); /*int [42]*/
e voila!
Proven wrong once again.
READ THE STANDARDS.
READ THE C++ STANDARDS.
"A consistent rule is followed for multidimensional arrays.
If E
is an
ndimensional
array of rank
i´ j´ . . . ´k, then E appearing in an expression is
converted
to a
pointer to an (n - 1 )dimensional
array
with rank j´ . . . ´k. If the * operator, either
explicitly or
implicitly as a result of subscripting, is applied to
this pointer, the result is the pointedto
(n - 1 )dimensional
array, which itself is immediately converted
into a pointer."
Irrelevant; what is relevant is the type of an object.
I rest my case.
Lets see how he tries to wangle out of this one
The inane ramblings of a demented freak.
So it's safe to assert you have created your own version f the
C++
language. And you refuse to accept the real C++ language, even
when
it's
presented to you in plain black'n'white.
GL troll , noob and idiot.
I hope you were lying about your age because a 57 year old
behaving
like you do is only explainable if you have just woken up from a
40
year coma.
Take your pills.
So you accept you are nothing more than a troll, noob and idiot.
Or do
you accept that you were/are wrong?
I accept neither.
You do not accept the text in the C++ standard as a rule for the
language?
Of course I do; now show me what I have supposedly written that
contradicts the Standard; from what I have seen you have done exactly
that on numerous occasions in the past few months. You seem to have a
habit of posting parts of the standard unrelated to a point under
discussion and of posting false claims of contradictions of the
Standard.
Ok lets start with just one of your statements in this post:
int (*pa)[42] = new int[42][42];
"pa" is a pointer to a 1D array of type int[42]; this is an
indisputable fact. The fact that what it points to is the
initial
element of a 2D array of type int[42][42] is irrelevant.
Wrong its a pointer to a 2d array according to the C++ standards.
(*pa) produces the pointedto (n-1) dimensional array. That is.... if pa
is the pointedto 2dim array(which it is) then derefenrencing it should
result in a 1dim array.
(*pa) produces the one dimensional array because pa is a pointer to a
one dimensional array (not a pointer to a two dimensional array), as I
said; ergo the only thing I am contradicting is your nonsense not the
standard.
