perlfaq4.1

视频监控网络部分的协议ddns,的模块的实现代码,请大家大胆指正.

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Using \f(CW\*(C`Bit::Vector\*(C'\fR:
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.Vb 4
\&        use Bit::Vector;
\&        $vec = Bit::Vector\->new(32);
\&        $vec\->Chunk_List_Store(3, split(//, reverse "33653337357"));
\&        $dec = $vec\->to_Dec();
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.IP "How do I convert from decimal to octal" 4
.IX Item "How do I convert from decimal to octal"
Using \f(CW\*(C`sprintf\*(C'\fR:
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.Vb 1
\&        $oct = sprintf("%o", 3735928559);
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Using \f(CW\*(C`Bit::Vector\*(C'\fR:
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.Vb 3
\&        use Bit::Vector;
\&        $vec = Bit::Vector\->new_Dec(32, \-559038737);
\&        $oct = reverse join(\*(Aq\*(Aq, $vec\->Chunk_List_Read(3));
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.IP "How do I convert from binary to decimal" 4
.IX Item "How do I convert from binary to decimal"
Perl 5.6 lets you write binary numbers directly with
the \f(CW\*(C`0b\*(C'\fR notation:
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.Vb 1
\&        $number = 0b10110110;
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Using \f(CW\*(C`oct\*(C'\fR:
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.Vb 2
\&        my $input = "10110110";
\&        $decimal = oct( "0b$input" );
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Using \f(CW\*(C`pack\*(C'\fR and \f(CW\*(C`ord\*(C'\fR:
.Sp
.Vb 1
\&        $decimal = ord(pack(\*(AqB8\*(Aq, \*(Aq10110110\*(Aq));
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Using \f(CW\*(C`pack\*(C'\fR and \f(CW\*(C`unpack\*(C'\fR for larger strings:
.Sp
.Vb 3
\&        $int = unpack("N", pack("B32",
\&        substr("0" x 32 . "11110101011011011111011101111", \-32)));
\&        $dec = sprintf("%d", $int);
\&
\&        # substr() is used to left pad a 32 character string with zeros.
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Using \f(CW\*(C`Bit::Vector\*(C'\fR:
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.Vb 2
\&        $vec = Bit::Vector\->new_Bin(32, "11011110101011011011111011101111");
\&        $dec = $vec\->to_Dec();
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.IP "How do I convert from decimal to binary" 4
.IX Item "How do I convert from decimal to binary"
Using \f(CW\*(C`sprintf\*(C'\fR (perl 5.6+):
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.Vb 1
\&        $bin = sprintf("%b", 3735928559);
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Using \f(CW\*(C`unpack\*(C'\fR:
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.Vb 1
\&        $bin = unpack("B*", pack("N", 3735928559));
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Using \f(CW\*(C`Bit::Vector\*(C'\fR:
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.Vb 3
\&        use Bit::Vector;
\&        $vec = Bit::Vector\->new_Dec(32, \-559038737);
\&        $bin = $vec\->to_Bin();
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The remaining transformations (e.g. hex \-> oct, bin \-> hex, etc.)
are left as an exercise to the inclined reader.
.Sh "Why doesn't & work the way I want it to?"
.IX Subsection "Why doesn't & work the way I want it to?"
The behavior of binary arithmetic operators depends on whether they're
used on numbers or strings.  The operators treat a string as a series
of bits and work with that (the string \f(CW"3"\fR is the bit pattern
\&\f(CW00110011\fR).  The operators work with the binary form of a number
(the number \f(CW3\fR is treated as the bit pattern \f(CW00000011\fR).
.PP
So, saying \f(CW\*(C`11 & 3\*(C'\fR performs the \*(L"and\*(R" operation on numbers (yielding
\&\f(CW3\fR).  Saying \f(CW"11" & "3"\fR performs the \*(L"and\*(R" operation on strings
(yielding \f(CW"1"\fR).
.PP
Most problems with \f(CW\*(C`&\*(C'\fR and \f(CW\*(C`|\*(C'\fR arise because the programmer thinks
they have a number but really it's a string.  The rest arise because
the programmer says:
.PP
.Vb 3
\&        if ("\e020\e020" & "\e101\e101") {
\&                # ...
\&                }
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.PP
but a string consisting of two null bytes (the result of \f(CW"\e020\e020"
& "\e101\e101"\fR) is not a false value in Perl.  You need:
.PP
.Vb 3
\&        if ( ("\e020\e020" & "\e101\e101") !~ /[^\e000]/) {
\&                # ...
\&                }
.Ve
.Sh "How do I multiply matrices?"
.IX Subsection "How do I multiply matrices?"
Use the Math::Matrix or Math::MatrixReal modules (available from \s-1CPAN\s0)
or the \s-1PDL\s0 extension (also available from \s-1CPAN\s0).
.Sh "How do I perform an operation on a series of integers?"
.IX Subsection "How do I perform an operation on a series of integers?"
To call a function on each element in an array, and collect the
results, use:
.PP
.Vb 1
\&        @results = map { my_func($_) } @array;
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.PP
For example:
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.Vb 1
\&        @triple = map { 3 * $_ } @single;
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.PP
To call a function on each element of an array, but ignore the
results:
.PP
.Vb 3
\&        foreach $iterator (@array) {
\&                some_func($iterator);
\&                }
.Ve
.PP
To call a function on each integer in a (small) range, you \fBcan\fR use:
.PP
.Vb 1
\&        @results = map { some_func($_) } (5 .. 25);
.Ve
.PP
but you should be aware that the \f(CW\*(C`..\*(C'\fR operator creates an array of
all integers in the range.  This can take a lot of memory for large
ranges.  Instead use:
.PP
.Vb 4
\&        @results = ();
\&        for ($i=5; $i < 500_005; $i++) {
\&                push(@results, some_func($i));
\&                }
.Ve
.PP
This situation has been fixed in Perl5.005. Use of \f(CW\*(C`..\*(C'\fR in a \f(CW\*(C`for\*(C'\fR
loop will iterate over the range, without creating the entire range.
.PP
.Vb 3
\&        for my $i (5 .. 500_005) {
\&                push(@results, some_func($i));
\&                }
.Ve
.PP
will not create a list of 500,000 integers.
.Sh "How can I output Roman numerals?"
.IX Subsection "How can I output Roman numerals?"
Get the http://www.cpan.org/modules/by\-module/Roman module.
.Sh "Why aren't my random numbers random?"
.IX Subsection "Why aren't my random numbers random?"
If you're using a version of Perl before 5.004, you must call \f(CW\*(C`srand\*(C'\fR
once at the start of your program to seed the random number generator.
.PP
.Vb 1
\&         BEGIN { srand() if $] < 5.004 }
.Ve
.PP
5.004 and later automatically call \f(CW\*(C`srand\*(C'\fR at the beginning.  Don't
call \f(CW\*(C`srand\*(C'\fR more than once\*(--you make your numbers less random,
rather than more.
.PP
Computers are good at being predictable and bad at being random
(despite appearances caused by bugs in your programs :\-).  see the
\&\fIrandom\fR article in the \*(L"Far More Than You Ever Wanted To Know\*(R"
collection in http://www.cpan.org/misc/olddoc/FMTEYEWTK.tgz , courtesy
of Tom Phoenix, talks more about this.  John von Neumann said, \*(L"Anyone
who attempts to generate random numbers by deterministic means is, of
course, living in a state of sin.\*(R"
.PP
If you want numbers that are more random than \f(CW\*(C`rand\*(C'\fR with \f(CW\*(C`srand\*(C'\fR
provides, you should also check out the \f(CW\*(C`Math::TrulyRandom\*(C'\fR module from
\&\s-1CPAN\s0.  It uses the imperfections in your system's timer to generate
random numbers, but this takes quite a while.  If you want a better
pseudorandom generator than comes with your operating system, look at
\&\*(L"Numerical Recipes in C\*(R" at http://www.nr.com/ .
.Sh "How do I get a random number between X and Y?"
.IX Subsection "How do I get a random number between X and Y?"
To get a random number between two values, you can use the \f(CW\*(C`rand()\*(C'\fR
builtin to get a random number between 0 and 1. From there, you shift
that into the range that you want.
.PP
\&\f(CW\*(C`rand($x)\*(C'\fR returns a number such that \f(CW\*(C`0 <= rand($x) < $x\*(C'\fR. Thus
what you want to have perl figure out is a random number in the range
from 0 to the difference between your \fIX\fR and \fIY\fR.
.PP
That is, to get a number between 10 and 15, inclusive, you want a
random number between 0 and 5 that you can then add to 10.
.PP
.Vb 1
\&        my $number = 10 + int rand( 15\-10+1 );
.Ve
.PP
Hence you derive the following simple function to abstract
that. It selects a random integer between the two given
integers (inclusive), For example: \f(CW\*(C`random_int_between(50,120)\*(C'\fR.
.PP
.Vb 7
\&        sub random_int_between {
\&                my($min, $max) = @_;
\&                # Assumes that the two arguments are integers themselves!
\&                return $min if $min == $max;
\&                ($min, $max) = ($max, $min)  if  $min > $max;
\&                return $min + int rand(1 + $max \- $min);
\&                }
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.SH "Data: Dates"
.IX Header "Data: Dates"
.Sh "How do I find the day or week of the year?"
.IX Subsection "How do I find the day or week of the year?"
The localtime function returns the day of the year.  Without an
argument localtime uses the current time.
.PP
.Vb 1
\&        $day_of_year = (localtime)[7];
.Ve
.PP
The \f(CW\*(C`POSIX\*(C'\fR module can also format a date as the day of the year or
week of the year.
.PP
.Vb 3
\&        use POSIX qw/strftime/;
\&        my $day_of_year  = strftime "%j", localtime;
\&        my $week_of_year = strftime "%W", localtime;
.Ve
.PP
To get the day of year for any date, use \f(CW\*(C`POSIX\*(C'\fR's \f(CW\*(C`mktime\*(C'\fR to get
a time in epoch seconds for the argument to localtime.
.PP
.Vb 3
\&        use POSIX qw/mktime strftime/;
\&        my $week_of_year = strftime "%W",
\&                localtime( mktime( 0, 0, 0, 18, 11, 87 ) );
.Ve
.PP
The \f(CW\*(C`Date::Calc\*(C'\fR module provides two functions to calculate these.
.PP
.Vb 3
\&        use Date::Calc;
\&        my $day_of_year  = Day_of_Year(  1987, 12, 18 );
\&        my $week_of_year = Week_of_Year( 1987, 12, 18 );
.Ve
.Sh "How do I find the current century or millennium?"
.IX Subsection "How do I find the current century or millennium?"
Use the following simple functions:
.PP
.Vb 3
\&        sub get_century    {
\&                return int((((localtime(shift || time))[5] + 1999))/100);
\&                }
\&
\&        sub get_millennium {
\&                return 1+int((((localtime(shift || time))[5] + 1899))/1000);
\&                }
.Ve
.PP
On some systems, the \f(CW\*(C`POSIX\*(C'\fR module's \f(CW\*(C`strftime()\*(C'\fR function has been
extended in a non-standard way to use a \f(CW%C\fR format, which they
sometimes claim is the \*(L"century\*(R". 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.
.Sh "How can I compare two dates and find the difference?"
.IX Subsection "How can I compare two dates and find the difference?"
(contributed by brian d foy)
.PP
You could just store all your dates as a number and then subtract.
Life isn't always that simple though. If you want to work with
formatted dates, the \f(CW\*(C`Date::Manip\*(C'\fR, \f(CW\*(C`Date::Calc\*(C'\fR, or \f(CW\*(C`DateTime\*(C'\fR
modules can help you.
.Sh "How can I take a string and turn it into epoch seconds?"
.IX Subsection "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 \f(CW\*(C`timelocal\*(C'\fR in the standard
\&\f(CW\*(C`Time::Local\*(C'\fR module.  Otherwise, you should look into the \f(CW\*(C`Date::Calc\*(C'\fR
and \f(CW\*(C`Date::Manip\*(C'\fR modules from \s-1CPAN\s0.
.Sh "How can I find the Julian Day?"
.IX Subsection "How can I find the Julian Day?"
(contributed by brian d foy and Dave Cross)
.PP
You can use the \f(CW\*(C`Time::JulianDay\*(C'\fR module available on \s-1CPAN\s0.  Ensure
that you really want to find a Julian day, though, as many people have
different ideas about Julian days.  See
http://www.hermetic.ch/cal_stud/jdn.htm for instance.
.PP
You can also try the \f(CW\*(C`DateTime\*(C'\fR module, which can convert a date/time
to a Julian Day.
.PP
.Vb 2
\&        $ perl \-MDateTime \-le\*(Aqprint DateTime\->today\->jd\*(Aq
\&        2453401.5
.Ve
.PP
Or the modified Julian Day
.PP
.Vb 2
\&        $ perl \-MDateTime \-le\*(Aqprint DateTime\->today\->mjd\*(Aq
\&        53401
.Ve
.PP
Or even the day of the year (which is what some people think of as a
Julian day)
.PP
.Vb 2
\&        $ perl \-MDateTime \-le\*(Aqprint DateTime\->today\->doy\*(Aq
\&        31
.Ve
.Sh "How do I find yesterday's date?"
.IX Subsection "How do I find yesterday's date?"
(contributed by brian d foy)
.PP
Use one of the Date modules. The \f(CW\*(C`DateTime\*(C'\fR module makes it simple, and
give you the same time of day, only the day before.
.PP
.Vb 1
\&        use DateTime;
\&
\&        my $yesterday = DateTime\->now\->subtract( days => 1 );
\&
\&        print "Yesterday was $yesterday\en";
.Ve
.PP
You can also use the \f(CW\*(C`Date::Calc\*(C'\fR module using its \f(CW\*(C`Today_and_Now\*(C'\fR
function.
.PP
.Vb 1