Once again Ian Collins is incorrect

Discussion in 'C++' started by Paul, Apr 25, 2011.

  1. Paul

    Paul Guest

    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:

    <quote>
    >> 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.
    >

    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.
    Paul, Apr 25, 2011
    #1
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  2. Paul

    Ian Collins Guest

    On 04/25/11 11:59 AM, Paul wrote:
    > 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:

    > On 04/25/11 11:46 AM, Paul wrote:
    >> "Ian Collins"<> wrote:


    >>> (*pparr) and arr have the same type. So whatever applies to one,

    applies to the other.

    >> I told you they had the same type


    >> Under no circumstances can dereferencing pparr access the array in

    the same way as arr

    I'll leave it at that.

    Sorry to impose yet another Paul thread on the group.

    --
    Ian Collins
    Ian Collins, Apr 25, 2011
    #2
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  3. Paul

    Paul Guest

    "Ian Collins" <> wrote in message
    news:...
    > On 04/25/11 11:59 AM, Paul wrote:
    >> 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*
    Paul, Apr 25, 2011
    #3
  4. Am 25.04.2011 11:54, schrieb Paul:
    >
    > "Ian Collins"<> wrote in message
    > news:...
    >> On 04/25/11 11:59 AM, Paul wrote:
    >>> 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
    Peter Remmers, Apr 25, 2011
    #4
  5. Paul

    SG Guest

    On 25 Apr., 11:54, Paul wrote:
    > Ian Collins wrote:
    > >> 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
    SG, Apr 26, 2011
    #5
  6. Paul

    Paul Guest

    "Peter Remmers" <> wrote in message
    news:4db583e0$0$6975$-online.net...
    > Am 25.04.2011 11:54, schrieb Paul:


    <quote>
    >> 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.
    >

    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, Apr 26, 2011
    #6
  7. Paul

    Paul Guest

    "SG" <> wrote in message
    news:...
    <quote>
    >> 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.
    >

    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, Apr 26, 2011
    #7
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