I have this strange problem with split in that it does not by default split into the default array.
Below is some toy code.
#!/usr/bin/perl
$A="A:B:C:D";
split (":",$A);
print $_[0];
This does not print anything. However if I explicitly split into the default array like
#!/usr/bin/perl
$A="A:B:C:D";
@_=split (":",$A);
print $_[0];
It's correctly prints A. My perl version is v5.22.1.
split does not go to @_ by default. @_ does not work like $_. It's only for arguments to a function. perlvar says:
Within a subroutine the array @_ contains the parameters passed to that subroutine. Inside a subroutine, @_ is the default array for the array operators pop and shift.
If you run your program with use strict and use warnings you'll see
Useless use of split in void context at
However, split does use $_ as its second argument (the string that is split up) if nothing is supplied. But you must always use the return value for something.
split in void context used to populate @_
You have to assign the split to an array:
use strict;
use warnings;
my $string = "A:B:C:D";
my @array = split(/:/, $string);
print $array[0] . "\n";
$a @a - it's bad to use single letter names anyway, but especially when it is $a or $b which are used by sort
@_is not "the default array".@_is the array that contains parameters to a subroutine. There is nothing in the Perl documentation that calls it the default array.