shift returns the first value of the array. When no array is defined it
operates on the default array @_
You will most likely see the line you mention as the first line of a
subroutine. E.g
&call_sub('John');
sub call_sub {
my $name = shift;
print "Name is $name\n";
}
What's happening here is that the value passed to the subroutine call_sub is
stored in a locally scoped default array for that subroutine. When you
perform a shift on the array it returns the data passed. This is stored in
the local $name scalar for use within the subroutine. Some other ways you
may see this written are
my $name = $_[0]; # Uses a direct route to the first element of the
array
my ($one, $two) = @_; # When passing more than one value to the subroutine,
assign them to two different scalars within the array
HTH
John
-----Original Message-----
From: Wim De Hul [mailto:[EMAIL PROTECTED]]
Sent: 20 December 2001 15:48
To: [EMAIL PROTECTED]
Subject: Shift question...
Hello guys ( and girls),
While I was reading a script, I saw the lines:
my $var = shift;
I thought that shift puts a variable in an array? What does this mean?
Thanks,
Wim.
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