Perl references (like the Perl language itself) are higher level than
their C counterparts. Pointers expose the memory address wheres
references (to my knowledge and at least not normally) do not. This
opens the door to pointer arithmetic and some of the black magic
possible with pointers (includ
From: yitzle <[EMAIL PROTECTED]>
> On 10/10/07, Jenda Krynicky <[EMAIL PROTECTED]> wrote:
> >
> > From: yitzle <[EMAIL PROTECTED]>
> > > The purpose of my message was to clarify the part of the documentation
> > that
> > > Kaushal asked about.
> >
> > I think you failed on that.
>
>
> I would t
Jenda Krynicky wrote:
From: yitzle <[EMAIL PROTECTED]>
The purpose of my message was to clarify the part of the documentation that
Kaushal asked about.
I think you failed on that.
I am aware that Perl has pointers/references, as I mentioned, but the
question is not /about/ pointers, but
On 10/10/07, Jenda Krynicky <[EMAIL PROTECTED]> wrote:
>
> And BTW ... you are aware of the fact that if you do
>
> int *b;
> *b = 5;
>
> you cause the program to crash, right?
> You did not start by assigning a variable to a reference to
> something, you assigned to the thing already referenced by
On 10/10/07, Jenda Krynicky <[EMAIL PROTECTED]> wrote:
>
> From: yitzle <[EMAIL PROTECTED]>
> > On 10/10/07, Jenda Krynicky <[EMAIL PROTECTED]> wrote:
> > > > Anyhow, the point is that Perl doesn't have those confusing weird
> > > "pointer"
> > > > stuff. $a and $c do not "point" to the same plac
From: yitzle <[EMAIL PROTECTED]>
> On 10/10/07, Jenda Krynicky <[EMAIL PROTECTED]> wrote:
> > > Anyhow, the point is that Perl doesn't have those confusing weird
> > "pointer"
> > > stuff. $a and $c do not "point" to the same place, the just got the same
> > > value. (Well, Perl /does/ have point
On 10/10/07, Jenda Krynicky <[EMAIL PROTECTED]> wrote:
>
> From: yitzle <[EMAIL PROTECTED]>
> > On 10/10/07, Kaushal Shriyan <[EMAIL PROTECTED]> wrote:
> > >
> > > Can you please explain me with a sample code. If I understand it
> correctly
> > > does the below code holds true for your explanation
It may be easier to understand outside of computer context and delving
into pointers/references/memory addresses. I'm sure there are some good
tried and true analogies that I can't think of so I'll try to come of up
with one that fits well into computers without getting too far from
reality or too
From: yitzle <[EMAIL PROTECTED]>
> On 10/10/07, Kaushal Shriyan <[EMAIL PROTECTED]> wrote:
> >
> > Can you please explain me with a sample code. If I understand it correctly
> > does the below code holds true for your explanation
>
> Lets put it this way.
> I the world of C/C++, there's something
From: Rob Dixon <[EMAIL PROTECTED]>
> Kaushal Shriyan wrote:
> > Hi,
> >
> > I am referring to http://www.gnulamp.com/perlscalars.html
> >
> > $a = $b; # Assign $b to $a
> >
> > Note that when Perl assigns a value with *$a = $b* it makes a copy of $b and
> > then assigns that to $a. Therefore th
On 10/10/07, Kaushal Shriyan <[EMAIL PROTECTED]> wrote:
>
> Can you please explain me with a sample code. If I understand it correctly
> does the below code holds true for your explanation
>
Lets put it this way.
I the world of C/C++, there's something called a pointer.
The syntax of Perl and C/
Kaushal Shriyan wrote:
Thanks Rob
All they're saying is that $a = $b doesn't tie $a and $b together in any
way, it just copies the value of $b to $a. If you change $b after that
it won't alter $a again.
Can you please explain me with a sample code. If I understand it
correctly does the bel
On 10/10/07, Rob Dixon <[EMAIL PROTECTED]> wrote:
>
> Kaushal Shriyan wrote:
> > Hi,
> >
> > I am referring to http://www.gnulamp.com/perlscalars.html
> >
> > $a = $b; # Assign $b to $a
> >
> > Note that when Perl assigns a value with *$a = $b* it makes a copy of $b
> and
> > then assigns that to $
Kaushal Shriyan wrote:
Hi,
I am referring to http://www.gnulamp.com/perlscalars.html
$a = $b; # Assign $b to $a
Note that when Perl assigns a value with *$a = $b* it makes a copy of $b and
then assigns that to $a. Therefore the next time you change $b it will not
alter $a.
I did not understan
14 matches
Mail list logo
