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Re: If statement not working

by choroba (Archbishop)
on Jun 25, 2020 at 17:21 UTC ( #11118541=note: print w/replies, xml ) Need Help??


in reply to If statement not working

Short answer: Use =~ instead of = to bind a regex against a variable.

Long answer: The $_ is the default topic. You don't need to mention it in many contexts, matching a regex being one of them. So, you can shorten the code to

if (/^[\$]\w/) { print "\nThat is a scalar data type\n"; } elsif (m/^[\@]\w/) { print "\nThat is an array data type\n"; } else { print "\nData type not found\n"; }

When you use =, it changes the contents of $_, so after the first condition, it contains the result of the regex match, not the original user input.

Also, remove the my from $_. Lexical $_ was deprecated and is not supported in modern Perl versions. Use local if you want to incorporate the code into a larger codebase.

Update:

Moreover, you can also use a hash from sigils to data types, no regex and if/else needed:

print "Enter a string and I will determine the data type: "; chomp(local $_ = <STDIN>); my %sigil2type = ('$' => 'scalar', '@' => 'array', '%' => 'hash', '*' => 'GLOB'); my $sigil = substr $_, 0, 1; my $type = $sigil2type{$sigil} || 'unknown'; say "\nThat is a $type data type.";

Update2: The $_ will be populated with the match result, not with the regex. Thanks AnomalousMonk.

map{substr$_->[0],$_->[1]||0,1}[\*||{},3],[[]],[ref qr-1,-,-1],[{}],[sub{}^*ARGV,3]

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