FAQ 4.3 Why isn't my octal data interpreted correctly?

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4.3: Why isn't my octal data interpreted correctly?

    (contributed by brian d foy)

    You're probably trying to convert a string to a number, which Perl only
    converts as a decimal number. When Perl converts a string to a number,
    it ignores leading spaces and zeroes, then assumes the rest of the
    digits are in base 10:

            my $string = '0644';

            print $string + 0;  # prints 644

            print $string + 44; # prints 688, certainly not octal!

    This problem usually involves one of the Perl built-ins that has the
    same name a Unix command that uses octal numbers as arguments on the
    command line. In this example, "chmod" on the command line knows that
    its first argument is octal because that's what it does:

            %prompt> chmod 644 file

    If you want to use the same literal digits (644) in Perl, you have to
    tell Perl to treat them as octal numbers either by prefixing the digits
    with a 0 or using "oct":

            chmod(     0644, $file);   # right, has leading zero
            chmod( oct(644), $file );  # also correct

    The problem comes in when you take your numbers from something that Perl
    thinks is a string, such as a command line argument in @ARGV:

            chmod( $ARGV[0],      $file);   # wrong, even if "0644"

            chmod( oct($ARGV[0]), $file );  # correct, treat string as octal

    You can always check the value you're using by printing it in octal
    notation to ensure it matches what you think it should be. Print it in
    octal and decimal format:

            printf "0%o %d", $number, $number;


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