Re: Can arrays be used as scalars?

[Paul Barker]

Hi

What your friend has done is assign the last 
element of the array @dir to $free. Perl array elements are accessed as 
$array_name[element_index] where the element index begins at zero. So to get the 
1st element of @dir you could write print $dir[0], to get the third $dir[2] etc 
...

The index of the last element can be found in $#dir 
so to get the last element you can write $dir[$#dir]. 

To find out how many elements there are in an array 
you could write $num_elemants = $#dir + 1 or you can just evaluate @dir in a 
scalar context ie print scalar(@dir). If you try print @dir however you will get 
the whole array as the print operator works in list context not scalar 
context.

If this is allreally confusing (sorry!) the 
check out thePerl docs (try 'perldoc perldata') or the Camel book, both 
ofwhich probably explain things a whole lot 
better than me !

HTH

Paul


  - Original Message - 
  From: 
  RCTay 
  
  To: [EMAIL PROTECTED] 
  
  Sent: Sunday, November 24, 2002 8:06 
  AM
  Subject: Can arrays be used as 
  scalars?
  
  my( @dir, 
  $free);@dir =3D `dir`;$free =3D $dir[$#dir];
  
  This code was 
  written by a friend of mine. "@dir" was declared as aarray, but in the 
  second line it is used as a scalar. How is thatpossible? There are no 
  warnings given by the interpreter. Can you pleaseexplain why it can be 
  used?

Re: Can arrays be used as scalars?

[csaba . raduly]

On 24/11/2002 08:06:35 RCTay wrote:

my( @dir,  $free);
@dir =3D `dir`;
$free =3D $dir[$#dir];

This code was  written by a friend of mine. @dir was declared as
a array, but
in the  second line it is used as a scalar. How is that possible? There
are no
warnings given by the interpreter. Can you please explain why it can be
used?

Neither the second nor the third line is valid Perl.

Bareword found where operator expected at -e line 1, near 3D
(Missing operator before D?)
Backticks found where operator expected at -e line 1, near D `dir`
(Do you need to predeclare D?)

Or did you mean something else ?

--
Csaba Ráduly, Software Engineer   Sophos Anti-Virus
email: [EMAIL PROTECTED]http://www.sophos.com
US Support: +1 888 SOPHOS 9 UK Support: +44 1235 559933


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Re: Can arrays be used as scalars?

[tran bao toan]

please do not send me anymore email, thanks!
[EMAIL PROTECTED] wrote:
On 24/11/2002 08:06:35 RCTay wrote:my( @dir, $free);@dir =3D `dir`;$free =3D $dir[$#dir];This code was written by a friend of mine. "@dir" was declared asaarray, butin the second line it is used as a scalar. How is thatpossible? Thereare nowarnings given by the interpreter. Can you pleaseexplain why it can beused?Neither the second nor the third line is valid Perl.Bareword found where operator expected at -e line 1, near "3D"(Missing operator before D?)Backticks found where operator expected at -e line 1, near "D `dir`"(Do you need to predeclare D?)Or did you mean something else ?--Csaba Ráduly, Software Engineer Sophos Anti-Virusemail: [EMAIL PROTECTED] http://www.sophos.comUS Support: +1 888 SOPHOS 9 UK Support: +44 1235 559933___Perl-Win32-Users mailing list[EMAIL PROTECTED]To unsubscribe: http://listserv.ActiveState.com/mailman/mysubsDo you Yahoo!?
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