Can't call method "print" on an undefined value - perl

Question

Can't call method print on an undefined value in line 40 line 2.
Here is the code. I use FileHandle to settle files:
#!/usr/bin/perl
use strict;
use warnings;
use FileHandle;
die unless (#ARGV ==4|| #ARGV ==5);
my #input =();
$input[0]=$ARGV[3];
$input[1]=$ARGV[4] if ($#ARGV==4);
chomp #input;
$input[0] =~ /([^\/]+)$/;
my $out = "$1.insert";
my $lane= "$1";
my %fh=();
open (Info,">$ARGV[1]") || die "$!";
open (AA,"<$ARGV[0]") || die "$!";
while(<AA>){
chomp;
my #inf=split;
my $iden=$inf[0];
my $outputfile="$ARGV[2]/$iden";
$fh{$iden}=FileHandle->new(">$outputfile");
}
close AA;
foreach my $input (#input) {
open (IN, "<$input" ) or die "$!" ;
my #path=split (/\//,$input);
print Info "#$path[-1]\n";
while (<IN>) {
my $line1 = $_;
my ($id1,$iden1) = (split "\t", $line1)[6,7];
my $line2 = <IN> ;
my ($id2,$iden2) = (split "\t", $line2)[6,7];
if ($id1 eq '+' && $id2 eq '-') {
my #inf=split(/\t/,$line1);
$fh{$iden1}->print($line1);
$fh{$iden2}->print($line2);
}
}
close IN;
}
I’ve tried multiple variations of this, but none of them seem to work. Any ideas?

Answer 1

Please remember that the primary worth of a Stack Overflow post is not to fix your particular problem, but to help the thousands of others who may be stuck in the same way. With that in mind, "I fixed it, thanks, bye" is more than a little selfish
As I said in my comment, using open directly on a hash element is much preferable to involving FileHandle. Perl will autovivify the hash element and create a file handle for you, and most people at all familiar with Perl will thank you for not making them read up again on the FileHandle documentation
I rewrote your code like this, which is much more Perlish and relies less on "magic numbers" to access #ARGV. You should really assign #ARGV to a list of named scalars, or - better still - use Getopt::Long so that they are named anyway
You should open your file handles as late as possible, and close the output handles early. This is effected most easily by using lexical file handles and limiting their scope to a block. Perl will implicitly close lexical handles for you when they go out of scope
There is no need to chomp the contents of #ARGVunless you could be be called under strange and errant circumstances, in which case you need to do a hell of a lot more to verify the input
You never use the result of $input[0] =~ /([^\/]+)$/ or the variables $out and $lane, so I removed them
#!/usr/bin/perl
use strict;
use warnings 'all';
# $ARGV[0] -- input file
# $ARGV[1] -- output log file
# $ARGV[2] -- directory for outputs per ident
# $ARGV[3] -- 1, $input[0]
# $ARGV[4] -- 2, $input[1] or undef
die "Fix the parameters" unless #ARGV == 4 or #ARGV == 5;
my #input = #ARGV[3,4];
my %fh;
{
open my $fh, '<', $ARGV[0] or die $!;
while ( <$fh> ) {
my $id = ( split )[0];
my $outputfile = "$ARGV[2]/$id";
open $fh{$id}, '>', $outputfile or die qq{Unable to open "$outputfile" for output: $!};
}
}
open my $log_fh, '>', $ARGV[1] or die qq{Unable to open "$ARGV[1]" for output: $!};
for my $input ( #input ) {
next unless $input; # skip unspecified parameters
my #path = split qr|/|, $input; # Really should be done by File::Spec
print $log_fh "#$path[-1]\n"; # Or File::Basename
open my $fh, '<', $input or die qq{Unable to open "$input" for input: $!};
while ( my $line0 = <$fh> ) {
chomp $line0;
my $line1 = <$fh>;
chomp $line1;
my ($id0, $iden0) = (split /\t/, $line0)[6,7];
my ($id1, $iden1) = (split /\t/, $line1)[6,7];
if ( $id0 eq '+' and $id1 eq '-' ) {
$fh{$_} or die qq{No output file for "$_"} for $iden0, $iden1;
print { $fh{$iden0} } $line0;
print { $fh{$iden1} } $line1;
}
}
}
while ( my ($iden, $fh) = each %fh ) {
close $fh or die qq{Unable to close file handle for "$iden": $!};
}

Answer 2

You don't have any error handling on this line:
$fh{$iden}=FileHandle->new(">$outputfile");
It's possible that opening a filehandle is silently failing, and only producing an error when you try to print to it. For example, if you have specified an invalid filename.
Also, you never check if $iden1 and $iden2 are names of open filehandles that actually exist. It's possible one of them does not exist.
In particular, you aren't removing a newline from $line1, so if $iden1 and $iden2 happen to be the last values on the line, this will be included in the name you are trying to use, and it will fail.

Answer 3

In your first while loop, you set up a hash of filehandles that you will write to later. The keys in this hash are the "iden" strings from the first file passed to the program.
Later, you parse another file and use the "iden" values in that file to choose which filehandle to write data to. But one (or more) of the "iden" values in the second file is missing from the first file. So that filehandle can't be found in the %fh hash. Because you don't check for that, you get `undef back from the hash and you can't print to an undefined filehandle.
To fix it, put a check before trying to use one of the filehandles from the %fh hash.
die "Unknown fh identifier '$iden1'" unless exists $fh{$iden1};
die "Unknown fh identifier '$iden2'" unless exists $fh{$iden2};
$fh{$iden1}->print($line1);
$fh{$iden2}->print($line2);