% perl -le '$foo=0.0; print $foo ? "True" : "False";'
False
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% perl -le '$foo="0.0"; print $foo ? "True" : "False";'
True
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Perl seems to be assuming that "0.0" is a string, which is fair enough, if that is by design, which I assume it is. "return 0.0" would from a subroutine would evaluate as false.

Perl will convert "0.0" to a number as soon as you do anything mathematical on it. Such as:

perl -e '$foo = "0.0" + 0; print $foo ? "True" : "False"'
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Which behaves as you'd expect. This reminds me of (even though it's completely different):


perl -e 'print "I cannot add" unless 19.08 + 2.01 == 21.09'
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If I remember correctly this comes from an early TPJ article.

Update: removed ' from print statement - thanks tilly - That's me in hybrid Win32 cmd.exe, Linux shell quoting mode :)

Simon Flack ($code or die)
$,=reverse'"ro_';s,$,\$,;s,$,lc ref sub{},e;$,
=~y'_"' ';eval"die";print $_,lc substr$@,0,3;

You need to do one of the following to make that work:

perl -e 'print "I cannot add" unless 19.08 + 2.01 == 21.09'
perl -e 'print "I can'\''t add" unless 19.08 + 2.01 == 21.09'
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BTW as Re: Strange (rounding?) problem points out, Perl often handles floats better if you use eq instead of == for testing equality...

Hey good point! "0.0" is only zero if you force an eval on it one way or another. These are interesting:

print "0.0 is == 0\n" if "0.0" == 0;
$string_zero = "0.0";
print "True" if $string_zero;
print "True" if eval $string_zero;
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cheers

tachyon

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