I think what is being overlooked is the effect of evaluating the empty statement block {}, which appears in all of the examples of the OP.

In the statement if ('') {}, the last expression evaluated is the conditional, which is false (the empty string); this is the value returned.

In the statement if ('true') {}, the last expression evaluated is the true clause, which is the empty statement block, which evaluates to the empty string; this is the value returned.

In the statement if (! 'true') {}, the last expression evaluated is the conditional, which evaluates to false (the empty string), etc.

This can be confirmed by re-writing the if statement to the 'statement modifier' form and using do {} for the empty statement (because {} on its own will be interpreted as an anonymous hash constructor).

E.g.: 

C:\@Work\Perl>perl -wMstrict -e "sub S { if (! $_[0]) {} }
print qq(<@{[ S($_) ]}> \n) for @ARGV"   ""  "hiya"
<>
<>

C:\@Work\Perl>perl -wMstrict -e "sub S { if (  $_[0]) {} }
print qq(<@{[ S($_) ]}> \n) for @ARGV"   ""  "hiya"
<>
<>

C:\@Work\Perl>perl -wMstrict -e "sub S { unless (! $_[0]) {} }
print qq(<@{[ S($_) ]}> \n) for @ARGV"   ""  "hiya"
<1>
<>

C:\@Work\Perl>perl -wMstrict -e "sub S { unless (  $_[0]) {} }
print qq(<@{[ S($_) ]}> \n) for @ARGV"   ""  "hiya"
<>
<hiya>

C:\@Work\Perl>perl -wMstrict -e "sub S { do {} if ! $_[0] }
print qq(<@{[ S($_) ]}> \n) for @ARGV"   ""  "hiya"
<>
<>

C:\@Work\Perl>perl -wMstrict -e "sub S { do {} if   $_[0] }
print qq(<@{[ S($_) ]}> \n) for @ARGV"   ""  "hiya"
<>
<>

C:\@Work\Perl>perl -wMstrict -e "sub S { do {} unless ! $_[0] }
print qq(<@{[ S($_) ]}> \n) for @ARGV"   ""  "hiya"
<1>
<>

C:\@Work\Perl>perl -wMstrict -e "sub S { do {} unless   $_[0] }
print qq(<@{[ S($_) ]}> \n) for @ARGV"   ""  "hiya"
<>
<hiya>

[download]

In reply to Re: unless versus if ( ! ) inside a subroutine by Anonymous Monk
in thread unless versus if ( ! ) inside a subroutine by Wheeler

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