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=head1 NAME

perlfaq4 - Data Manipulation ($Revision: 10394 $)

=head1 DESCRIPTION

This section of the FAQ answers questions related to manipulating
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)?

Internally, your computer represents floating-point numbers in binary.
Digital (as in powers of two) computers cannot store all numbers
exactly.  Some real numbers lose precision in the process.  This is a
problem with how computers store numbers and affects all computer
languages, not just Perl.

L<perlnumber> shows the gory details of number representations and
conversions.

To limit the number of decimal places in your numbers, you can use the
printf or sprintf function.  See the L<"Floating Point
Arithmetic"|perlop> for more details.

	printf "%.2f", 10/3;

	my $number = sprintf "%.2f", 10/3;

=head2 Why is int() broken?

Your C<int()> is most probably working just fine.  It's the numbers that
aren't quite what you think.

First, see the answer to "Why am I getting long decimals
(eg, 19.9499999999999) instead of the numbers I should be getting
(eg, 19.95)?".

For example, this

	print int(0.6/0.2-2), "\n";

will in most computers print 0, not 1, because even such simple
numbers as 0.6 and 0.2 cannot be presented exactly by floating-point
numbers.  What you think in the above as 'three' is really more like
2.9999999999999995559.

=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.  Octal literals in perl must start with a
leading C<0> and hexadecimal literals must start with a leading C<0x>.
If they are read in from somewhere and assigned, no automatic
conversion takes place.  You must explicitly use C<oct()> or C<hex()> if you
want the values converted to decimal.  C<oct()> interprets hexadecimal (C<0x350>),
octal (C<0350> or even without the leading C<0>, like C<377>) and binary
(C<0b1010>) numbers, while C<hex()> only converts hexadecimal ones, with
or without a leading C<0x>, such as C<0x255>, C<3A>, C<ff>, or C<deadbeef>.
The inverse mapping from decimal to octal can be done with either the
<%o> or C<%O> C<sprintf()> formats.

This problem shows up most often when people try using C<chmod()>,
C<mkdir()>, C<umask()>, or C<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
C<644>, rather than the intended octal literal C<0644>.  The problem can
be seen with:

	printf("%#o",644);   # prints 01204

Surely you had not intended C<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 C<0..7>.

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

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

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

The C<POSIX> module (part of the standard Perl distribution)
implements C<ceil()>, C<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 C<Math::Complex>
module.  With 5.004, the C<Math::Trig> module (part of the standard Perl
distribution) implements the trigonometric functions. Internally it
uses the C<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 between numeric representations/bases/radixes?

As always with Perl there is more than one way to do it.  Below are a
few examples of approaches to making common conversions between number
representations.  This is intended to be representational rather than
exhaustive.

Some of the examples later in L<perlfaq4> use the C<Bit::Vector>
module from CPAN. The reason you might choose C<Bit::Vector> over the
perl built in functions is that it works with numbers of ANY size,
that it is optimized for speed on some operations, and for at least
some programmers the notation might be familiar.

=over 4

=item How do I convert hexadecimal into decimal

Using perl's built in conversion of C<0x> notation:

	$dec = 0xDEADBEEF;

Using the C<hex> function:

	$dec = hex("DEADBEEF");

Using C<pack>:

	$dec = unpack("N", pack("H8", substr("0" x 8 . "DEADBEEF", -8)));

Using the CPAN module C<Bit::Vector>:

	use Bit::Vector;
	$vec = Bit::Vector->new_Hex(32, "DEADBEEF");
	$dec = $vec->to_Dec();

=item How do I convert from decimal to hexadecimal

Using C<sprintf>:

	$hex = sprintf("%X", 3735928559); # upper case A-F
	$hex = sprintf("%x", 3735928559); # lower case a-f

Using C<unpack>:

	$hex = unpack("H*", pack("N", 3735928559));

Using C<Bit::Vector>:

	use Bit::Vector;
	$vec = Bit::Vector->new_Dec(32, -559038737);
	$hex = $vec->to_Hex();

And C<Bit::Vector> supports odd bit counts:

	use Bit::Vector;
	$vec = Bit::Vector->new_Dec(33, 3735928559);
	$vec->Resize(32); # suppress leading 0 if unwanted
	$hex = $vec->to_Hex();

=item How do I convert from octal to decimal

Using Perl's built in conversion of numbers with leading zeros:

	$dec = 033653337357; # note the leading 0!

Using the C<oct> function:

	$dec = oct("33653337357");

Using C<Bit::Vector>:

	use Bit::Vector;
	$vec = Bit::Vector->new(32);
	$vec->Chunk_List_Store(3, split(//, reverse "33653337357"));
	$dec = $vec->to_Dec();

=item How do I convert from decimal to octal

Using C<sprintf>:

	$oct = sprintf("%o", 3735928559);

Using C<Bit::Vector>:

	use Bit::Vector;
	$vec = Bit::Vector->new_Dec(32, -559038737);
	$oct = reverse join('', $vec->Chunk_List_Read(3));

=item How do I convert from binary to decimal

Perl 5.6 lets you write binary numbers directly with
the C<0b> notation:

	$number = 0b10110110;

Using C<oct>:

	my $input = "10110110";
	$decimal = oct( "0b$input" );

Using C<pack> and C<ord>:

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

Using C<pack> and C<unpack> for larger strings:

	$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.

Using C<Bit::Vector>:

	$vec = Bit::Vector->new_Bin(32, "11011110101011011011111011101111");
	$dec = $vec->to_Dec();

=item How do I convert from decimal to binary

Using C<sprintf> (perl 5.6+):

	$bin = sprintf("%b", 3735928559);

Using C<unpack>:

	$bin = unpack("B*", pack("N", 3735928559));

Using C<Bit::Vector>:

	use Bit::Vector;
	$vec = Bit::Vector->new_Dec(32, -559038737);
	$bin = $vec->to_Bin();

The remaining transformations (e.g. hex -> oct, bin -> hex, etc.)
are left as an exercise to the inclined reader.

=back

=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<3>).  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.cpan.org/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.

	 BEGIN { srand() if $] < 5.004 }

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 :-).  see the
F<random> article in the "Far More Than You Ever Wanted To Know"
collection in http://www.cpan.org/misc/olddoc/FMTEYEWTK.tgz , 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 C<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/ .

=head2 How do I get a random number between X and Y?

To get a random number between two values, you can use the C<rand()>
builtin to get a random number between 0 and 1. From there, you shift
that into the range that you want.

C<rand($x)> returns a number such that C<< 0 <= rand($x) < $x >>. Thus
what you want to have perl figure out is a random number in the range
from 0 to the difference between your I<X> and I<Y>.

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.

	my $number = 10 + int rand( 15-10+1 );

Hence you derive the following simple function to abstract
that. It selects a random integer between the two given
integers (inclusive), For example: C<random_int_between(50,120)>.

	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);
		}

=head1 Data: Dates

=head2 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.

	$day_of_year = (localtime)[7];

The C<POSIX> module can also format a date as the day of the year or
week of the year.

	use POSIX qw/strftime/;
	my $day_of_year  = strftime "%j", localtime;
	my $week_of_year = strftime "%W", localtime;

To get the day of year for any date, use C<POSIX>'s C<mktime> to get