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Next: do shell script "system_profiler...
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External
Since: Jan 28, 2004
Posts: 4
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(Msg. 1) Posted: Wed Jan 28, 2004 6:16 pm
Post subject: Is this a known Bug?
Archived from groups: alt>comp>lang>applescript (more info?)
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In the following code listA changes after listB is modified with a
repeat loop. Why does that happen? If listB is changed outside a repeat
loop, listA is unchanged.
set listA to {{1, 1}, {2, 2}, {3, 3}, {4, 4}}
set listB to listA
repeat with itemx in listB
set item 2 of itemx to (item 1 of itemx) + 1
end repeat
return {listA, listB}
-->{{{1, 2}, {2, 3}, {3, 4}, {4, 5}},
-->{{1, 2}, {2, 3}, {3, 4}, {4, 5}}}
Thanks,
Greg Robb |
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External
Since: Jan 28, 2004
Posts: 4
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(Msg. 2) Posted: Wed Jan 28, 2004 7:43 pm
Post subject: Re: Is this a known Bug? [Login to view extended thread Info.]
Archived from groups: per prev. post (more info?)
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In article <280120040729332982%greg_robb@home.net>, greg robb
<greg_robb.TakeThisOut@home.net> wrote:
Oops! ListA also changes when listB is modified outside of a repeat
loop. Why is listA modified when listB is changed?
set listA to {{1, 1}, {2, 2}, {3, 3}, {4, 4}}
set listB to listA
set item 1 of listB to {1, 7}
return {listA, listB}
-->{{{1, 7}, {2, 2}, {3, 3}, {4, 4}},
-->{{1, 7}, {2, 2}, {3, 3}, {4, 4}}}
> In the following code listA changes after listB is modified with a
> repeat loop. Why does that happen? If listB is changed outside a repeat
> loop, listA is unchanged.
>
>
> set listA to {{1, 1}, {2, 2}, {3, 3}, {4, 4}}
> set listB to listA
> repeat with itemx in listB
> set item 2 of itemx to (item 1 of itemx) + 1
> end repeat
>
> return {listA, listB}
> -->{{{1, 2}, {2, 3}, {3, 4}, {4, 5}},
> -->{{1, 2}, {2, 3}, {3, 4}, {4, 5}}}
>
> Thanks,
> Greg Robb<!-- ~MESSAGE_AFTER~ --> |
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External
Since: Nov 12, 2003
Posts: 360
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(Msg. 3) Posted: Wed Jan 28, 2004 9:00 pm
Post subject: Re: Is this a known Bug? [Login to view extended thread Info.]
Archived from groups: per prev. post (more info?)
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In article <280120040856220852%greg_robb@home.net>,
Greg Robb <greg_robb RemoveThis @home.net> wrote:
> In article <280120040729332982%greg_robb@home.net>, greg robb
> <greg_robb RemoveThis @home.net> wrote:
>
>
> Oops! ListA also changes when listB is modified outside of a repeat
> loop. Why is listA modified when listB is changed?
Because the "set" command, when used with a list, does not make a copy
of the data. It simply makes listB points to the same data as listA. If
you want a duplicate copy, you need to use "copy listA to listB".
See : <http://developer.apple.com/documentation/AppleScript/Conceptual/
AppleScriptLangGuide/index.html>
Patrick
--
Patrick Stadelmann <Patrick.Stadelmann RemoveThis @unine.ch><!-- ~MESSAGE_AFTER~ --> >> Stay informed about: Is this a known Bug? |
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External
Since: Nov 12, 2003
Posts: 360
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(Msg. 4) Posted: Wed Jan 28, 2004 9:03 pm
Post subject: Re: Is this a known Bug? [Login to view extended thread Info.]
Archived from groups: per prev. post (more info?)
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In article
<Patrick.Stadelmann-ABCD86.18002528012004 RemoveThis @news.fu-berlin.de>,
Patrick Stadelmann <Patrick.Stadelmann RemoveThis @unine.ch> wrote:
> In article <280120040856220852%greg_robb@home.net>,
> Greg Robb <greg_robb RemoveThis @home.net> wrote:
>
> > In article <280120040729332982%greg_robb@home.net>, greg robb
> > <greg_robb RemoveThis @home.net> wrote:
> >
> >
> > Oops! ListA also changes when listB is modified outside of a repeat
> > loop. Why is listA modified when listB is changed?
>
> Because the "set" command, when used with a list, does not make a copy
> of the data. It simply makes listB points to the same data as listA. If
> you want a duplicate copy, you need to use "copy listA to listB".
>
> See : <http://developer.apple.com/documentation/AppleScript/Conceptual/
> AppleScriptLangGuide/index.html>
Oops, here's the full URL : <http://developer.apple.com/documentation/
AppleScript/Conceptual/AppleScriptLangGuide/AppleScript.9a.html>
Patrick
--
Patrick Stadelmann <Patrick.Stadelmann RemoveThis @unine.ch><!-- ~MESSAGE_AFTER~ --> >> Stay informed about: Is this a known Bug? |
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External
Since: Jan 28, 2004
Posts: 1
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(Msg. 5) Posted: Wed Jan 28, 2004 9:03 pm
Post subject: Re: Is this a known Bug? [Login to view extended thread Info.]
Archived from groups: per prev. post (more info?)
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Thanks Patrick, I thought I was going crazy.
So using "set" shares the data between the two variables as long as a
concatenation or deleting an item doesn't happen.
set listA to {{1, 1}, {2, 2}, {3, 3}, {4, 4}}
set listB to listA
set listB to items 1 thru 3 of listB
return {listA, listB}
-->{{{1, 1}, {2, 2}, {3, 3}, {4, 4}},
-->{{1, 1}, {2, 2}, {3, 3}}}
set listA to {{1, 1}, {2, 2}, {3, 3}, {4, 4}}
set listB to listA
set listB to listB & {7, 7}
return {listA, listB}
-->{{{1, 1}, {2, 2}, {3, 3}, {4, 4}},
-->{{1, 1}, {2, 2}, {3, 3}, {4, 4}, 7, 7}}
set listA to {{1, 1}, {2, 2}, {3, 3}, {4, 4}}
set listB to listA
set listB to items 1 thru 3 of listB & {{8, 8}}
return {listA, listB}
-->{{{1, 1}, {2, 2}, {3, 3}, {4, 4}},
-->{{1, 1}, {2, 2}, {3, 3}, {8, 8}}}
-Greg Robb >> Stay informed about: Is this a known Bug? |
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External
Since: Jul 09, 2003
Posts: 925
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(Msg. 6) Posted: Wed Jan 28, 2004 10:47 pm
Post subject: Re: Is this a known Bug? [Login to view extended thread Info.]
Archived from groups: per prev. post (more info?)
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In article <e50b567e.0401281622.4e3a2de0.TakeThisOut@posting.google.com>, Greg Robb
<grobb.TakeThisOut@agtnet.com> wrote:
> Thanks Patrick, I thought I was going crazy.
>
> So using "set" shares the data between the two variables as long as a
> concatenation or deleting an item doesn't happen.
It only happens with lists, also, not with simple data types.
--
Jerry Kindall, Seattle, WA <http://www.jerrykindall.com/>
Send only plain text messages under 32K to the Reply-To address.
This mailbox is filtered aggressively to thwart spam and viruses.<!-- ~MESSAGE_AFTER~ --> >> Stay informed about: Is this a known Bug? |
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External
Since: Jan 26, 2004
Posts: 34
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(Msg. 7) Posted: Fri Jan 30, 2004 1:39 pm
Post subject: Re: Is this a known Bug? [Login to view extended thread Info.]
Archived from groups: per prev. post (more info?)
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On 28/01/2004, Greg Robb wrote in message
<e50b567e.0401281622.4e3a2de0.TakeThisOut@posting.google.com>:
> So using "set" shares the data between the two variables as long as a
> concatenation or deleting an item doesn't happen.
>
> set listA to {{1, 1}, {2, 2}, {3, 3}, {4, 4}}
> set listB to listA
> set listB to items 1 thru 3 of listB
> return {listA, listB}
> -->{{{1, 1}, {2, 2}, {3, 3}, {4, 4}},
> -->{{1, 1}, {2, 2}, {3, 3}}}
Correct -- as long as data isn't actually changed. It might help to
think of the 'set' command as automatically having 'a reference to'
in it unless you specify otherwise.
Simon.
--
Posted using test version of software.
Please tell me if anything isn't right.<!-- ~MESSAGE_AFTER~ --> >> Stay informed about: Is this a known Bug? |
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