Mathias Pasquay wrote:
> Dear list,
Hello,
> i'am just wondering about the behaviour of shift and the defaultvariable @_ .
>
> I use the following code::
>
> while (<RULEFILE>) { # each line of RULEFILE is stored in
> $_
> chomp; # delete \n from $_
> split /;/; # split the fields of $_ seperated
> by ; into @_
You should have gotten the warning "Use of implicit split to @_ is
deprecated" which means that you shouldn't do that. You should enable
warnings.
my @array = split /;/;
> print "Array before shift: @_\n"; # prints @_
> my $test = shift @_; # $test gets the first
> item from @_
> print "Array after shift: @_\n"; # print @_ without the
> first item
> }
>
> everything works fine.
>
> Now i change the line "my $test = shift @_" into "my $test = shift". This
> should work because shift should use @_ by default.
No, in this case it uses @ARGV by default.
perldoc -f shift
> But in this case $test is empty an the second print command prints still the
> whole @_.
>
> I really don't know why this should not work?
>
> I would be happy about any help.
John
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