On Nov 15, Jennifer Garner said:
I can't understand for this script below,I want somebody here to give me
some help.Thanks.
sub is_tainted{
my $var=shift;
my $blank=substr($var,0,0);
return not eval {eval "1 || $blank" || 1};
}
That subroutine estimate for if some given var is tainted or not.But I
can't know how it works.
If a variable is tainted, then any substring of that variable is also
tainted. In addition, it is illegal to eval() any string that is tainted.
The inclusion of a tainted string inside another string makes that whole
string tainted.
Therefore: if $var is tainted, the $blank will also be tainted (even
though it's a substring of zero characters). If $blank is then tainted,
then the code
eval { eval "1 || $blank" || 1 }
will return false (since the eval { ... } catches fatal errors, and the
eval "1 || $blank" raises a fatal error because $blank is tainted), and
therefore,
return not eval { eval "1 || $blank" || 1 };
returns true, stating that $var is indeed tainted. If $var wasn't
tainted, then
eval "1 || $blank" || 1
returns 1, and
return not eval { 1 }
returns false, stating that $var wasn't tainted.
*whew*
Frankly, I find the 'eval "1 || $blank" || 1' silly, since the whole
reason the '... || 1' is needed is since $blank is a blank string and the
code '1 || ' is invalid Perl. Long story short, I'd have written:
sub is_tainted {
return not eval { eval 1 . substr($_[0], 0, 0) };
}
It's much more concise. If $_[0] isn't tainted, then
not eval { eval 1 . substr($_[0], 0, 0) }
->
not eval { eval 1 }
->
not eval { 1 }
->
not 1
->
false
whereas if $_[1] is tainted, then the eval { ... } returns false since a
fatal error is raised because
eval 1 . substr($_[0], 0, 0)
is illegal if $_[0] is tainted.
--
Jeff "japhy" Pinyan % How can we ever be the sold short or
RPI Acacia Brother #734 % the cheated, we who for every service
http://www.perlmonks.org/ % have long ago been overpaid?
http://princeton.pm.org/ % -- Meister Eckhart
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