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PROVEN WRONG!

 
 
Paul
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      03-18-2011

"Leigh Johnston" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed) ...
> On 18/03/2011 17:06, Paul wrote:
>>
>> "Leigh Johnston" <(E-Mail Removed)> wrote in message
>> news:(E-Mail Removed) ...
>>> On 18/03/2011 16:44, Paul wrote:
>>>>
>>>> "Leigh Johnston" <(E-Mail Removed)> wrote in message
>>>> news:(E-Mail Removed) ...
>>>>> On 18/03/2011 16:16, Paul wrote:
>>>>>>
>>>>>> "Leigh Johnston" <(E-Mail Removed)> wrote in message
>>>>>> news:(E-Mail Removed)...
>>>>>>> On 18/03/2011 13:53, Paul wrote:
>>>>>>>>
>>>>
>>>> <snip>
>>>>> I hope you are a troll; I really do; if you aren't a troll and you
>>>>> genuinely believe the garbage you post then your situation is probably
>>>>> hopeless.
>>>>>
>>>>> /Leigh
>>>>>
>>>>
>>>> Leigh you have nothing sensible to say because I have once again PROVEN
>>>> YOU WRONG.
>>>> Now why don't you shut up and quietly go away and accept that instead
>>>> of
>>>> behaving like a 6 year old child who had their sweeties taken .
>>>>
>>>
>>> The way you repeatedly describe yourself when trying to insult others
>>> is quite uncanny.
>>>
>>>>>>> Maybe the following will help you understand the semantic
>>>>>>> differences
>>>>>>> involved:
>>>>>>>
>>>>>>> int main()
>>>>>>> {
>>>>>>> int* p1 = new int[42]; // 'p1' is a pointer to the first element
>>>>>>> of an
>>>>>>> array
>>>>>>> int (*p2)[42] = new int[1][42]; // 'p2' is a pointer to an array
>>>>>>> }
>>>>>>>
>>>>>> This is a clear indication of how screwed up you are. Please read the
>>>>>> following :
>>>>>>
>>>>>> "the newexpression yields a pointer to the initial element (if any)
>>>>>> of the
>>>>>> array. [Note: both new int and new int[10] have type int* and the
>>>>>> type
>>>>>> of new int[i][10] is
>>>>>> int (*)[10]. ]"
>>>>>>
>>>>>>
>>>>>> The standards define both of these types to be pointers to the array,
>>>>>> they are just different *types* of pointer.
>>>>>>

>> PROVEN WRONG !!

>
> What have I been proven wrong about exactly? Where is this proof? BTW
> "pointer to the initial element of an array" is not the same as "pointer
> to an array".
>
>> I have nothing more to say , shut up and go away child.

>
> Look in the mirror if you want to see something childish.
>
>>>>> Maybe the following will help you understand the semantic differences
>>>>> involved:
>>>>>
>>>>> int main()
>>>>> {
>>>>> int* p1 = new int[42]; // 'p1' is a pointer to the first element of an
>>>>> array
>>>>> int (*p2)[42] = new int[1][42]; // 'p2' is a pointer to an array
>>>>> }
>>>>>
>>>> This is a clear indication of how screwed up you are. Please read the
>>>> following :
>>>>
>>>> "the newexpression yields a pointer to the initial element (if any)
>>>> of the
>>>> array. [Note: both new int and new int[10] have type int* and the type
>>>> of new int[i][10] is
>>>> int (*)[10]. ]"
>>>>
>>>>
>>>> The standards define both of these types to be pointers to the array,
>>>> they are just different *types* of pointer.
>>>>

PROVEN WRONG !!
I have nothing more to say .


 
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Paul
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      03-18-2011

"Leigh Johnston" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed) ...
> On 18/03/2011 17:40, Paul wrote:
>>
>> "Leigh Johnston" <(E-Mail Removed)> wrote in message
>> news:(E-Mail Removed) ...
>>> On 18/03/2011 17:06, Paul wrote:
>>>>
>>>> "Leigh Johnston" <(E-Mail Removed)> wrote in message
>>>> news:(E-Mail Removed) ...
>>>>> On 18/03/2011 16:44, Paul wrote:
>>>>>>
>>>>>> "Leigh Johnston" <(E-Mail Removed)> wrote in message
>>>>>> news:(E-Mail Removed) ...
>>>>>>> On 18/03/2011 16:16, Paul wrote:
>>>>>>>>
>>>>>>>> "Leigh Johnston" <(E-Mail Removed)> wrote in message
>>>>>>>> news:(E-Mail Removed)...
>>>>>>>>> On 18/03/2011 13:53, Paul wrote:
>>>>>>>>>>
>>>>>>
>>>>>> <snip>
>>>>>>> I hope you are a troll; I really do; if you aren't a troll and you
>>>>>>> genuinely believe the garbage you post then your situation is
>>>>>>> probably
>>>>>>> hopeless.
>>>>>>>
>>>>>>> /Leigh
>>>>>>>
>>>>>>
>>>>>> Leigh you have nothing sensible to say because I have once again
>>>>>> PROVEN
>>>>>> YOU WRONG.
>>>>>> Now why don't you shut up and quietly go away and accept that
>>>>>> instead of
>>>>>> behaving like a 6 year old child who had their sweeties taken .
>>>>>>
>>>>>
>>>>> The way you repeatedly describe yourself when trying to insult others
>>>>> is quite uncanny.
>>>>>
>>>>>>>>> Maybe the following will help you understand the semantic
>>>>>>>>> differences
>>>>>>>>> involved:
>>>>>>>>>
>>>>>>>>> int main()
>>>>>>>>> {
>>>>>>>>> int* p1 = new int[42]; // 'p1' is a pointer to the first element
>>>>>>>>> of an
>>>>>>>>> array
>>>>>>>>> int (*p2)[42] = new int[1][42]; // 'p2' is a pointer to an array
>>>>>>>>> }
>>>>>>>>>
>>>>>>>> This is a clear indication of how screwed up you are. Please read
>>>>>>>> the
>>>>>>>> following :
>>>>>>>>
>>>>>>>> "the newexpression yields a pointer to the initial element (if any)
>>>>>>>> of the
>>>>>>>> array. [Note: both new int and new int[10] have type int* and the
>>>>>>>> type
>>>>>>>> of new int[i][10] is
>>>>>>>> int (*)[10]. ]"
>>>>>>>>
>>>>>>>>
>>>>>>>> The standards define both of these types to be pointers to the
>>>>>>>> array,
>>>>>>>> they are just different *types* of pointer.
>>>>>>>>
>>>> PROVEN WRONG !!
>>>
>>> What have I been proven wrong about exactly? Where is this proof? BTW
>>> "pointer to the initial element of an array" is not the same as
>>> "pointer to an array".
>>>
>>>> I have nothing more to say , shut up and go away child.
>>>
>>> Look in the mirror if you want to see something childish.
>>>
>>>>>>> Maybe the following will help you understand the semantic
>>>>>>> differences
>>>>>>> involved:
>>>>>>>
>>>>>>> int main()
>>>>>>> {
>>>>>>> int* p1 = new int[42]; // 'p1' is a pointer to the first element
>>>>>>> of an
>>>>>>> array
>>>>>>> int (*p2)[42] = new int[1][42]; // 'p2' is a pointer to an array
>>>>>>> }
>>>>>>>
>>>>>> This is a clear indication of how screwed up you are. Please read the
>>>>>> following :
>>>>>>
>>>>>> "the newexpression yields a pointer to the initial element (if any)
>>>>>> of the
>>>>>> array. [Note: both new int and new int[10] have type int* and the
>>>>>> type
>>>>>> of new int[i][10] is
>>>>>> int (*)[10]. ]"
>>>>>>
>>>>>>
>>>>>> The standards define both of these types to be pointers to the array,
>>>>>> they are just different *types* of pointer.

>
> The Standard defines the type of the first new-expression in that example
> to be a pointer to a scalar type (int) and the second to be a pointer to
> an array.
>


The standard defines that both types to point to the initial elelment of the
array, the only difference is that one is a multi-dimesional array and one
is a single dimension.
The new expression returns a pointer to the first element of the array ,
that is a pointer to the array, regardless of what type it is.

You still do not understand the implications of E1[E2] being identical to
*((E1)+(E2)). This means the E1 will always be a pointer of type int* for a
1dimensional array of int. Whether its implicitly converted in case of
static arrays, or doesn't need to be converted for dynamic arrays.

int (*E1)[16] = new int[3][16];
int* E1 = new int[16];

E1 in the first example is no more a pointer to the array than in the second
example, both of these pointer types point to an array, and both of them are
dereferenced to an object of type exactly:
ArrayDim-1, for example:
If E1 was a 3Dim array dereferncing E1 would produce a 2Dim array
If E1 was a 2Dim array dereferencing it would produce a 1Dim array
If E1 was a 1Dim array , dereferencing it would produce the base object
type.


ALthough I can't help feeling ima wasting my time speaking to you because
you repeatedly reply with idiotic nonsense and utter bullshit. No doubt the
same sort of reply will be seen to this post.




>>>>>>

>> PROVEN WRONG !!

>


 
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Paul
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Posts: n/a
 
      03-18-2011

"Leigh Johnston" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed) ...
> On 18/03/2011 18:36, Paul wrote:
>>
>>>>>>>> <snip>
>>>>>>>>> I hope you are a troll; I really do; if you aren't a troll and you
>>>>>>>>> genuinely believe the garbage you post then your situation is
>>>>>>>>> probably
>>>>>>>>> hopeless.
>>>>>>>>>
>>>>>>>>> /Leigh
>>>>>>>>>
>>>>>>>>
>>>>>>>> Leigh you have nothing sensible to say because I have once again
>>>>>>>> PROVEN
>>>>>>>> YOU WRONG.
>>>>>>>> Now why don't you shut up and quietly go away and accept that
>>>>>>>> instead of
>>>>>>>> behaving like a 6 year old child who had their sweeties taken .
>>>>>>>>
>>>>>>>
>>>>>>> The way you repeatedly describe yourself when trying to insult
>>>>>>> others
>>>>>>> is quite uncanny.
>>>>>>>
>>>>>>>>>>> Maybe the following will help you understand the semantic
>>>>>>>>>>> differences
>>>>>>>>>>> involved:
>>>>>>>>>>>
>>>>>>>>>>> int main()
>>>>>>>>>>> {
>>>>>>>>>>> int* p1 = new int[42]; // 'p1' is a pointer to the first element
>>>>>>>>>>> of an
>>>>>>>>>>> array
>>>>>>>>>>> int (*p2)[42] = new int[1][42]; // 'p2' is a pointer to an array
>>>>>>>>>>> }
>>>>>>>>>>>
>>>>>>>>>> This is a clear indication of how screwed up you are. Please read
>>>>>>>>>> the
>>>>>>>>>> following :
>>>>>>>>>>
>>>>>>>>>> "the newexpression yields a pointer to the initial element (if
>>>>>>>>>> any)
>>>>>>>>>> of the
>>>>>>>>>> array. [Note: both new int and new int[10] have type int* and the
>>>>>>>>>> type
>>>>>>>>>> of new int[i][10] is
>>>>>>>>>> int (*)[10]. ]"
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> The standards define both of these types to be pointers to the
>>>>>>>>>> array,
>>>>>>>>>> they are just different *types* of pointer.
>>>>>>>>>>
>>>>>> PROVEN WRONG !!
>>>>>
>>>>> What have I been proven wrong about exactly? Where is this proof? BTW
>>>>> "pointer to the initial element of an array" is not the same as
>>>>> "pointer to an array".
>>>>>
>>>>>> I have nothing more to say , shut up and go away child.
>>>>>
>>>>> Look in the mirror if you want to see something childish.
>>>>>
>>>>>>>>> Maybe the following will help you understand the semantic
>>>>>>>>> differences
>>>>>>>>> involved:
>>>>>>>>>
>>>>>>>>> int main()
>>>>>>>>> {
>>>>>>>>> int* p1 = new int[42]; // 'p1' is a pointer to the first element
>>>>>>>>> of an
>>>>>>>>> array
>>>>>>>>> int (*p2)[42] = new int[1][42]; // 'p2' is a pointer to an array
>>>>>>>>> }
>>>>>>>>>
>>>>>>>> This is a clear indication of how screwed up you are. Please read
>>>>>>>> the
>>>>>>>> following :
>>>>>>>>
>>>>>>>> "the newexpression yields a pointer to the initial element (if any)
>>>>>>>> of the
>>>>>>>> array. [Note: both new int and new int[10] have type int* and the
>>>>>>>> type
>>>>>>>> of new int[i][10] is
>>>>>>>> int (*)[10]. ]"
>>>>>>>>
>>>>>>>>
>>>>>>>> The standards define both of these types to be pointers to the
>>>>>>>> array,
>>>>>>>> they are just different *types* of pointer.
>>>
>>> The Standard defines the type of the first new-expression in that
>>> example to be a pointer to a scalar type (int) and the second to be a
>>> pointer to an array.
>>>

>>
>> The standard defines that both types to point to the initial elelment of
>> the array, the only difference is that one is a multi-dimesional array
>> and one is a single dimension.
>> The new expression returns a pointer to the first element of the array ,
>> that is a pointer to the array, regardless of what type it is.

>
> The first element of an array is not the same thing as the array which
> contains the element. int* is not a pointer to an array, int* is a
> pointer to a scalar type (int).
>


It *IS* a pointer the the array. How can it point to any part of the array
yet not point to the array?
As I predicited a completely idiotic nonsensical response.


>>
>> You still do not understand the implications of E1[E2] being identical
>> to *((E1)+(E2)). This means the E1 will always be a pointer of type int*
>> for a 1dimensional array of int. Whether its implicitly converted in
>> case of static arrays, or doesn't need to be converted for dynamic
>> arrays.

>
> No it doesn't. An array is not a pointer and a pointer is not an array;
> an array can be converted to a pointer however.
>

The standard agrees with me not you ref:
"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."

Once agian you are proven wrong.



>>
>> int (*E1)[16] = new int[3][16];
>> int* E1 = new int[16];
>>
>> E1 in the first example is no more a pointer to the array than in the
>> second example, both of these pointer types point to an array, and both
>> of them are dereferenced to an object of type exactly

>
> E1 is a pointer to an array in the first example; in the second example it
> is not a pointer to an array as it is a pointer to a scalar type (int).

Look another idiotic response ^^
>
> :
>> ArrayDim-1, for example:
>> If E1 was a 3Dim array dereferncing E1 would produce a 2Dim array
>> If E1 was a 2Dim array dereferencing it would produce a 1Dim array
>> If E1 was a 1Dim array , dereferencing it would produce the base object
>> type.

>
> Correct; well almost; each dereference produces a reference of the
> corresponding array element type. The problem here is not your
> understanding of how arrays work at the most basic level; the problem here
> is the language you have been using to express your understanding; the
> language you have been using has been plain incorrect.
>

I know it's correct I'm telling you how it works.
>>
>>
>> ALthough I can't help feeling ima wasting my time speaking to you
>> because you repeatedly reply with idiotic nonsense and utter bullshit.
>> No doubt the same sort of reply will be seen to this post.

>
> You are failing to acknowledge that the language you use to express
> technical C++ matters is irking all those who reply to your posts. Your
> insults will not help you improve your understanding of technical matters
> nor help you express yourself properly.
>

As I predicited mostly idiotic responses , no wonder I simply get ****ed off
with this idiot half the time and simply tell him to f off for lack of a
better expression.

 
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