Re^4: Looking for pointers or optimizations.

[thmsdrew]

in reply to Re^3: Looking for pointers or optimizations.
in thread Looking for pointers or optimizations.

So if I put in while (<$fh>) it breaks my my $guess = <> further down in my code. I made that my $guess = <STDIN> and it works fine.

[2teez]

So if I put in while (<$fh>) it breaks my my $guess = <> further down in my code. I made that my $guess = <STDIN> and it works fine.

both while(<>){..} and $guess = <> reads from STDIN. Atleast, that is obvious from your code. So, to make while(<fh>){.. work, you have to do this:


my ($words_file) = @ARGV;
open( my $fh, "<", $words_file ) or die "can't open file: $!";
my @words;
while (<$fh>) {
  ...
}

## later you use
chomp(my $guess = <STDIN>);
...

Please, pay attention to the DIRECTION of the "arrow"

• ">" is 'Writing to' ( OUTPUT )
• "<" is 'Read from' ( INPUT )
NOTE: Like I mentioned previously, you didn't close the your filehandle!!!

[SerZKO]

Hej, You probably didn't see this (marked with ####)

open(my $fh, "<", $ARGV[0]) or die "cannot open input file: $!";
my @words = <$fh>;
close $fh or die "cannot close input file: $!"; ##############
my $words = @words;

On the other hand, I don't understand the last line - you read a file into array with my @words = <$fh>; and then you assign it to a scalar variablemy $words = @words;.

OK, you probably have your own reasons for that but you could manipulate array as good as scalar. And you could read a file directly in a scalar variable. my $words = <$fh>;

[choroba]

you could read a file directly in a scalar variable.

my $words = <$fh>;

Not true. Have you tried it? In fact, you can, but in a different way.

[chromatic]

And you could read a file directly in a scalar...

Depending on the current value of the input record separator, you might only get one record. In fact, you'll probably only get one record.