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comp.lang.ruby

Array variable.

Andreu

11/7/2008 3:57:00 PM

Please consider the following code:

def myfunct

ie = 0
0.step(il, 3) do |i|
dmy = ilines[i+1]
icatn[ie] = dmy[2,5]
idate[ie] = dmy[18,5]
ie += 1
end

me = 0
0.step(ml, 3) do |j|
dmy = mlines[j+1]
mcatn[me] = dmy[2,5]
mdate[me] = dmy[18,5]
me += 1
end

end


When the program exits the first loop, the contents of array idate is OK.
When the second loop is exited, the array variable idate have the same
contents as mdate ...

Any clue welcomed, thanks, Andreu.
6 Answers

Brian Adkins

11/7/2008 4:31:00 PM

0

Andreu <root@sys1.org> writes:

> Please consider the following code:
>
> def myfunct
>
> ie = 0
> 0.step(il, 3) do |i|
> dmy = ilines[i+1]
> icatn[ie] = dmy[2,5]
> idate[ie] = dmy[18,5]
> ie += 1
> end
>
> me = 0
> 0.step(ml, 3) do |j|
> dmy = mlines[j+1]
> mcatn[me] = dmy[2,5]
> mdate[me] = dmy[18,5]
> me += 1
> end
>
> end
>
>
> When the program exits the first loop, the contents of array idate is OK.
> When the second loop is exited, the array variable idate have the same
> contents as mdate ...
>
> Any clue welcomed, thanks, Andreu.

You've left out some important info, but aren't you assigning the same
values to both arrays?

idate[ie] = dmy[18,5]
....
mdate[me] = dmy[18,5]

--
Brian Adkins
http://www....
http://lojic...

Andreu

11/7/2008 7:20:00 PM

0

Brian,
Thanks for your answer.
No, if you look again at the code, you will see that in the first loop
'dmy' is read from an array named 'ilines' and in the second one
is read from another array named 'mlines' so should be different.

Andreu.

Brian Adkins wrote:
> Andreu <root@sys1.org> writes:
>
>
> idate[ie] = dmy[18,5]
> ....
> mdate[me] = dmy[18,5]
>

Andreu

11/7/2008 10:33:00 PM

0

PROBLEM SOLVED
At the beginning of my program I have written this line:

idate = mdate = Array.new

but it seems Ruby doesn't support this syntax (like C) and assume
that the two arrays are the same one.

Thanks for your time, Andreu.

Simon Krahnke

11/11/2008 2:32:00 AM

0

* Andreu <root@sys1.org> (2008-11-07) schrieb:

> PROBLEM SOLVED
> At the beginning of my program I have written this line:
>
> idate = mdate = Array.new
>
> but it seems Ruby doesn't support this syntax (like C) and assume
> that the two arrays are the same one.

Actually Ruby and C do the same thing here: Assign whatever Array.new
yields to mdate and idate.

mfg, simon .... l

Charles Calvert

11/11/2008 9:50:00 PM

0

On Fri, 07 Nov 2008 23:33:12 +0100, Andreu <root@sys1.org> wrote in
<gf2flo$te8$1@aioe.org>:

>PROBLEM SOLVED
>At the beginning of my program I have written this line:
>
>idate = mdate = Array.new
>
>but it seems Ruby doesn't support this syntax (like C) and assume
>that the two arrays are the same one.

It's not a difference in syntax, but a difference in how variables are
handled. When you write something like this in C:

int a, b;

a = b = 0;

You are creating two variables on the stack and assigning an integer
to them. The actual value 0 is stored in borth variables because they
are intrinsic types that have been allocated on the stack.

When you write in Ruby:

idate = mdate = Array.new

something different happens. Array.new returns a reference to an
instance of array that has been allocated on the heap. The variable
mdate holds the value of the reference, not the instance itself. This
same reference is then assigned to idate. Now both variables have a
reference to the same instace on the heap.

A reference is really just a pointer with nicer semantics. If you're
familiar with a language that has pointers, like C, then think of it
like this:

int * GetInt()
{
int * i = (int *) malloc(sizeof(int));
return i;
}

int * i, * j;

i = j = GetInt();

Now if you modify j:

*j = 5;
printf("%d", *i); // prints "5"
--
Charles Calvert | Web-site Design/Development
Celtic Wolf, Inc. | Software Design/Development
http://www.celti... | Data Conversion
(703) 580-0210 | Project Management

Andreu

11/13/2008 2:11:00 PM

0

Ok, it's clear now. Thanks for your good description on what's going on.

Andreu.



Charles Calvert wrote:
>
> It's not a difference in syntax, but a difference in how variables are
> handled. When you write something like this in C: