perlfaq4.pod

MSYS在windows下模拟了一个类unix的终端

=head1 NAME

perlfaq4 - Data Manipulation ($Revision: 1.49 $, $Date: 1999/05/23 20:37:49 $)

=head1 DESCRIPTION

The section of the FAQ answers questions related to the manipulation
of data as numbers, dates, strings, arrays, hashes, and miscellaneous
data issues.

=head1 Data: Numbers

=head2 Why am I getting long decimals (eg, 19.9499999999999) instead of the numbers I should be getting (eg, 19.95)?

The infinite set that a mathematician thinks of as the real numbers can
only be approximated on a computer, since the computer only has a finite
number of bits to store an infinite number of, um, numbers.

Internally, your computer represents floating-point numbers in binary.
Floating-point numbers read in from a file or appearing as literals
in your program are converted from their decimal floating-point
representation (eg, 19.95) to an internal binary representation.

However, 19.95 can't be precisely represented as a binary
floating-point number, just like 1/3 can't be exactly represented as a
decimal floating-point number.  The computer's binary representation
of 19.95, therefore, isn't exactly 19.95.

When a floating-point number gets printed, the binary floating-point
representation is converted back to decimal.  These decimal numbers
are displayed in either the format you specify with printf(), or the
current output format for numbers.  (See L<perlvar/"$#"> if you use
print.  C<$#> has a different default value in Perl5 than it did in
Perl4.  Changing C<$#> yourself is deprecated.)

This affects B<all> computer languages that represent decimal
floating-point numbers in binary, not just Perl.  Perl provides
arbitrary-precision decimal numbers with the Math::BigFloat module
(part of the standard Perl distribution), but mathematical operations
are consequently slower.

To get rid of the superfluous digits, just use a format (eg,
C<printf("%.2f", 19.95)>) to get the required precision.
See L<perlop/"Floating-point Arithmetic">.  

=head2 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.  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.  oct() interprets
both hex ("0x350") numbers and octal ones ("0350" or even without the
leading "0", like "377"), while hex() only converts hexadecimal ones,
with or without a leading "0x", like "0x255", "3A", "ff", or "deadbeef".

This problem shows up most often when people try using chmod(), mkdir(),
umask(), or sysopen(), which all want permissions in octal.

    chmod(644,  $file);	# WRONG -- perl -w catches this
    chmod(0644, $file);	# right

=head2 Does Perl have a round() function?  What about ceil() and floor()?  Trig functions?

Remember that int() merely truncates toward 0.  For rounding to a
certain number of digits, sprintf() or printf() is usually the easiest
route.

    printf("%.3f", 3.1415926535);	# prints 3.142

The POSIX module (part of the standard Perl distribution) implements
ceil(), floor(), and a number of other mathematical and trigonometric
functions.

    use POSIX;
    $ceil   = ceil(3.5);			# 4
    $floor  = floor(3.5);			# 3

In 5.000 to 5.003 perls, trigonometry was done in the Math::Complex
module.  With 5.004, the Math::Trig module (part of the standard Perl
distribution) implements the trigonometric functions. Internally it
uses the Math::Complex module and some functions can break out from
the real axis into the complex plane, for example the inverse sine of
2.

Rounding in financial applications can have serious implications, and
the rounding method used should be specified precisely.  In these
cases, it probably pays not to trust whichever system rounding is
being used by Perl, but to instead implement the rounding function you
need yourself.

To see why, notice how you'll still have an issue on half-way-point
alternation:

    for ($i = 0; $i < 1.01; $i += 0.05) { printf "%.1f ",$i}

    0.0 0.1 0.1 0.2 0.2 0.2 0.3 0.3 0.4 0.4 0.5 0.5 0.6 0.7 0.7 
    0.8 0.8 0.9 0.9 1.0 1.0

Don't blame Perl.  It's the same as in C.  IEEE says we have to do this.
Perl numbers whose absolute values are integers under 2**31 (on 32 bit
machines) will work pretty much like mathematical integers.  Other numbers
are not guaranteed.

=head2 How do I convert bits into ints?

To turn a string of 1s and 0s like C<10110110> into a scalar containing
its binary value, use the pack() and unpack() functions (documented in
L<perlfunc/"pack"> and L<perlfunc/"unpack">):

    $decimal = unpack('c', pack('B8', '10110110'));

This packs the string C<10110110> into an eight bit binary structure.
This is then unpacked as a character, which returns its ordinal value.

This does the same thing:

    $decimal = ord(pack('B8', '10110110'));

Here's an example of going the other way:

    $binary_string = unpack('B*', "\x29");

=head2 Why doesn't & work the way I want it to?

The behavior of binary arithmetic operators depends on whether they're
used on numbers or strings.  The operators treat a string as a series
of bits and work with that (the string C<"3"> is the bit pattern
C<00110011>).  The operators work with the binary form of a number
(the number C<3> is treated as the bit pattern C<00000011>).

So, saying C<11 & 3> performs the "and" operation on numbers (yielding
C<1>).  Saying C<"11" & "3"> performs the "and" operation on strings
(yielding C<"1">).

Most problems with C<&> and C<|> arise because the programmer thinks
they have a number but really it's a string.  The rest arise because
the programmer says:

    if ("\020\020" & "\101\101") {
	# ...
    }

but a string consisting of two null bytes (the result of C<"\020\020"
& "\101\101">) is not a false value in Perl.  You need:

    if ( ("\020\020" & "\101\101") !~ /[^\000]/) {
	# ...
    }

=head2 How do I multiply matrices?

Use the Math::Matrix or Math::MatrixReal modules (available from CPAN)
or the PDL extension (also available from CPAN).

=head2 How do I perform an operation on a series of integers?

To call a function on each element in an array, and collect the
results, use:

    @results = map { my_func($_) } @array;

For example:

    @triple = map { 3 * $_ } @single;

To call a function on each element of an array, but ignore the
results:

    foreach $iterator (@array) {
        some_func($iterator);
    }

To call a function on each integer in a (small) range, you B<can> use:

    @results = map { some_func($_) } (5 .. 25);

but you should be aware that the C<..> operator creates an array of
all integers in the range.  This can take a lot of memory for large
ranges.  Instead use:

    @results = ();
    for ($i=5; $i < 500_005; $i++) {
        push(@results, some_func($i));
    }

This situation has been fixed in Perl5.005. Use of C<..> in a C<for>
loop will iterate over the range, without creating the entire range.

    for my $i (5 .. 500_005) {
        push(@results, some_func($i));
    }

will not create a list of 500,000 integers.

=head2 How can I output Roman numerals?

Get the http://www.perl.com/CPAN/modules/by-module/Roman module.

=head2 Why aren't my random numbers random?

If you're using a version of Perl before 5.004, you must call C<srand>
once at the start of your program to seed the random number generator.
5.004 and later automatically call C<srand> at the beginning.  Don't
call C<srand> more than once--you make your numbers less random, rather
than more.

Computers are good at being predictable and bad at being random
(despite appearances caused by bugs in your programs :-).
http://www.perl.com/CPAN/doc/FMTEYEWTK/random , courtesy of Tom
Phoenix, talks more about this.  John von Neumann said, ``Anyone who
attempts to generate random numbers by deterministic means is, of
course, living in a state of sin.''

If you want numbers that are more random than C<rand> with C<srand>
provides, you should also check out the Math::TrulyRandom module from
CPAN.  It uses the imperfections in your system's timer to generate
random numbers, but this takes quite a while.  If you want a better
pseudorandom generator than comes with your operating system, look at
``Numerical Recipes in C'' at http://www.nr.com/ .

=head1 Data: Dates

=head2 How do I find the week-of-the-year/day-of-the-year?

The day of the year is in the array returned by localtime() (see
L<perlfunc/"localtime">):

    $day_of_year = (localtime(time()))[7];

or more legibly (in 5.004 or higher):

    use Time::localtime;
    $day_of_year = localtime(time())->yday;

You can find the week of the year by dividing this by 7:

    $week_of_year = int($day_of_year / 7);

Of course, this believes that weeks start at zero.  The Date::Calc
module from CPAN has a lot of date calculation functions, including
day of the year, week of the year, and so on.   Note that not
all businesses consider ``week 1'' to be the same; for example,
American businesses often consider the first week with a Monday
in it to be Work Week #1, despite ISO 8601, which considers
WW1 to be the first week with a Thursday in it.

=head2 How do I find the current century or millennium?

Use the following simple functions:

    sub get_century    { 
	return int((((localtime(shift || time))[5] + 1999))/100);
    } 
    sub get_millennium { 
	return 1+int((((localtime(shift || time))[5] + 1899))/1000);
    } 

On some systems, you'll find that the POSIX module's strftime() function
has been extended in a non-standard way to use a C<%C> format, which they
sometimes claim is the "century".  It isn't, because on most such systems,
this is only the first two digits of the four-digit year, and thus cannot
be used to reliably determine the current century or millennium.

=head2 How can I compare two dates and find the difference?

If you're storing your dates as epoch seconds then simply subtract one
from the other.  If you've got a structured date (distinct year, day,
month, hour, minute, seconds values), then for reasons of accessibility,
simplicity, and efficiency, merely use either timelocal or timegm (from
the Time::Local module in the standard distribution) to reduce structured
dates to epoch seconds.  However, if you don't know the precise format of
your dates, then you should probably use either of the Date::Manip and
Date::Calc modules from CPAN before you go hacking up your own parsing
routine to handle arbitrary date formats.

=head2 How can I take a string and turn it into epoch seconds?

If it's a regular enough string that it always has the same format,
you can split it up and pass the parts to C<timelocal> in the standard
Time::Local module.  Otherwise, you should look into the Date::Calc
and Date::Manip modules from CPAN.

=head2 How can I find the Julian Day?

Use the Time::JulianDay module (part of the Time-modules bundle
available from CPAN.)

Before you immerse yourself too deeply in this, be sure to verify that it
is the I<Julian> Day you really want.  Are you really just interested in
a way of getting serial days so that they can do date arithmetic?  If you
are interested in performing date arithmetic, this can be done using
either Date::Manip or Date::Calc, without converting to Julian Day first.

There is too much confusion on this issue to cover in this FAQ, but the
term is applied (correctly) to a calendar now supplanted by the Gregorian
Calendar, with the Julian Calendar failing to adjust properly for leap
years on centennial years (among other annoyances).  The term is also used
(incorrectly) to mean: [1] days in the Gregorian Calendar; and [2] days
since a particular starting time or `epoch', usually 1970 in the Unix
world and 1980 in the MS-DOS/Windows world.  If you find that it is not
the first meaning that you really want, then check out the Date::Manip
and Date::Calc modules.  (Thanks to David Cassell for most of this text.)

=head2 How do I find yesterday's date?

The C<time()> function returns the current time in seconds since the
epoch.  Take twenty-four hours off that:

    $yesterday = time() - ( 24 * 60 * 60 );

Then you can pass this to C<localtime()> and get the individual year,
month, day, hour, minute, seconds values.

Note very carefully that the code above assumes that your days are
twenty-four hours each.  For most people, there are two days a year
when they aren't: the switch to and from summer time throws this off.
A solution to this issue is offered by Russ Allbery.

    sub yesterday {
	my $now  = defined $_[0] ? $_[0] : time;
	my $then = $now - 60 * 60 * 24;
	my $ndst = (localtime $now)[8] > 0;
	my $tdst = (localtime $then)[8] > 0;
	$then - ($tdst - $ndst) * 60 * 60;
    }
    # Should give you "this time yesterday" in seconds since epoch relative to
    # the first argument or the current time if no argument is given and
    # suitable for passing to localtime or whatever else you need to do with
    # it.  $ndst is whether we're currently in daylight savings time; $tdst is
    # whether the point 24 hours ago was in daylight savings time.  If $tdst
    # and $ndst are the same, a boundary wasn't crossed, and the correction
    # will subtract 0.  If $tdst is 1 and $ndst is 0, subtract an hour more
    # from yesterday's time since we gained an extra hour while going off
    # daylight savings time.  If $tdst is 0 and $ndst is 1, subtract a
    # negative hour (add an hour) to yesterday's time since we lost an hour.
    #
    # All of this is because during those days when one switches off or onto
    # DST, a "day" isn't 24 hours long; it's either 23 or 25.
    #
    # The explicit settings of $ndst and $tdst are necessary because localtime
    # only says it returns the system tm struct, and the system tm struct at
    # least on Solaris doesn't guarantee any particular positive value (like,
    # say, 1) for isdst, just a positive value.  And that value can
    # potentially be negative, if DST information isn't available (this sub
    # just treats those cases like no DST).
    #
    # Note that between 2am and 3am on the day after the time zone switches
    # off daylight savings time, the exact hour of "yesterday" corresponding
    # to the current hour is not clearly defined.  Note also that if used
    # between 2am and 3am the day after the change to daylight savings time,
    # the result will be between 3am and 4am of the previous day; it's
    # arguable whether this is correct.
    #
    # This sub does not attempt to deal with leap seconds (most things don't).
    #
    # Copyright relinquished 1999 by Russ Allbery <[e-mail]>
    # This code is in the public domain

=head2 Does Perl have a Year 2000 problem?  Is Perl Y2K compliant?

Short answer: No, Perl does not have a Year 2000 problem.  Yes, Perl is
Y2K compliant (whatever that means).  The programmers you've hired to
use it, however, probably are not.

Long answer: The question belies a true understanding of the issue.