FAQ 4.3 Why isn't my octal data interpreted correctly?
- From: PerlFAQ Server <comdog@xxxxxxxxx>
- Date: Fri, 29 Apr 2005 16:03:01 +0000 (UTC)
This message is one of several periodic postings to comp.lang.perl.misc
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common questions. The core of this message represents an excerpt
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4.3: Why isn't my octal data interpreted correctly?
Perl only understands octal and hex numbers as such when they occur as
literals in your program. Octal literals in perl must start with a
leading "0" and hexadecimal literals must start with a leading "0x". If
they are read in from somewhere and assigned, no automatic conversion
takes place. You must explicitly use oct() or hex() if you want the
values converted to decimal. oct() interprets hex ("0x350"), octal
("0350" or even without the leading "0", like "377") and binary
("0b1010") numbers, while hex() only converts hexadecimal ones, with or
without a leading "0x", like "0x255", "3A", "ff", or "deadbeef". The
inverse mapping from decimal to octal can be done with either the "%o"
or "%O" sprintf() formats.
This problem shows up most often when people try using chmod(), mkdir(),
umask(), or sysopen(), which by widespread tradition typically take
permissions in octal.
chmod(644, $file); # WRONG
chmod(0644, $file); # right
Note the mistake in the first line was specifying the decimal literal
644, rather than the intended octal literal 0644. The problem can be
seen with:
printf("%#o",644); # prints 01204
Surely you had not intended "chmod(01204, $file);" - did you? If you
want to use numeric literals as arguments to chmod() et al. then please
try to express them as octal constants, that is with a leading zero and
with the following digits restricted to the set 0..7.
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