I had a bug in my code which was a result of insufficient validation of a string passed to the built-in hex function.

In my opinion, hex is much too forgiving. As shown in its documentation, it allows for a string to have a leading 0x; for example, it converts 0xC to 12. Although x is not a legal hexadecimal character, it is customary to denote a hex value with the 0xC prefix. This is quite reasonable.

Although not mentioned in the documentation, it also allows for a string to have a leading x; for example, it converts xC to 12. I guess that's reasonable, too.

My problem occurred when I inadvertently passed a lone x to hex. The function returned 0. Even with warnings enabled, I did not get a warning message (perl v5.12.2, linux):

perl -w -E "say hex(q{x})"
0
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Passing it an illegal hex string like t does generate a warning, as desired:

perl -w -E "say hex(q{t})"
Illegal hexadecimal digit 't' ignored at -e line 1.
0
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I think passing a lone x should generate a warning message (with warnings enabled), but I'm not sure it's worth submitting a perlbug since I really need to do some checking before passing a string to hex anyway.

Here is a wrapper function I decided to use. It calls hex after performing some input validation:

sub hex2 {
    my $str = shift;
    $str =~ s/^(0x|x)//;
    if (length $str) {
        if ($str =~ /([^0-9a-f])/i) {
            die "hex2: Illegal hexadecimal digit found: '$1'";
        }
        else {
            return hex $str;
        }
    }
    else {
        die "hex2: No chars found after stripping leading 0x or x";
    }
}
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Another approach is to override hex.