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Fejezet 13. F�ggv�nyekFelhaszn�l� �ltal defini�lt f�ggv�nyek
F�ggv�nyeket a k�vetkez� szintaxis szerint defini�lhatod:
B�rmely �rv�nyes PHP k�d megjelenhet egy f�ggv�nyen bel�l,
ak�r m�g m�s f�ggv�ny vagy
oszt�ly
defin�ci�k is.
PHP 3-ban a f�ggv�nyeket defini�lni kell, miel�tt hivatkoz�s t�rt�nik r�juk
(f�ggv�nyh�v�s el�tt).
PHP 4-ben nincs ez a megk�t�s.
A PHP nem t�mogatja a f�ggv�nyek polimorfizmus�t (t�bbalak�s�g�t), a
f�ggv�nyekdefin�ci�kat nem lehet megsz�ntetni vagy �jradefini�lni egy
m�r defini�lt f�ggv�nyeket.
A PHP 3 nem t�mogatja a v�ltoz� sz�m� f�ggv�nyargumentumokat, b�r
az argumentumok kezd��rt�ke t�mogatott. L�sd az
Argumentumok kezd��rt�ke
c�m� r�szt b�vebb inform�ci��rt.
A PHP 4 mindkett� lehet�s�get t�mogatja. L�sd
a V�ltoz� sz�m� f�ggv�nyargumentumok
c�m� r�szt �s a func_num_args(), func_get_arg() �s a
func_get_args() f�ggv�nyeket r�szletesebb le�r�s�rt.
User Contributed Notes
F�ggv�nyek |
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[email protected]
08-Nov-1999 06:21 |
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Many people ask how to call functions that are in other files. See
"Require" and "Include" in the manual. Also the value
of $DOCUMENT_ROOT is good for including sub-includes. (It is NOT like
C/C++ includes. The dir is ALWAYS relative to the main source file.)
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[email protected]
05-Feb-2000 01:32 |
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Important Note to All New Users: functions do NOT have default access to
GLOBAL variables. You must specify globals as such in your function using
the 'global' type/keyword. See the section on
variables:scope.
This note should also be added to the
documentation, as it would help the majority of programmers who use
languages where globals are, well, global (that is, available from
anywhere). The scoping rules should also not be buried in subsection 4 of
the variables section. It should be front and center because I think this
is probably one of the most non-standard and thus confusing design choices
of PHP.
[Ed. note: the variables $_GET, $_POST, $_REQUEST,
$_SESSION, and $_FILES are superglobals, which means you don't need the
global keyword to use them inside a function]
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[email protected]
22-Feb-2000 10:19 |
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When using a function within a function, using global in the inner function
will not make variables available that have been first initialized within
the outer function.
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[email protected]
24-Apr-2000 08:02 |
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Stack overflow means your function called itself recursivly too many times
and just completely filled up the processes stack. That error is there to
stop a recursive call from completely taking up the entire system memory.
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[email protected]
05-Dec-2000 10:51 |
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If a user-defined function does not explicitly return a value, then it
will return NULL. For most truth tests, this can be considered as
FALSE.
For example:
function print_list
($array)
{
print implode ("<br />",
$array);
}
if (print_list ($HTTP_POST_VARS))
print
"The print_list function returned a value that can be considered
true."
else
print "The print_list function returned a
value that can be considered false."
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[email protected]
14-Dec-2000 09:14 |
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See also about controlling the generation of error messages by putting @ in
front of the function before you call it, in the section "error
control operators".
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21-Dec-2000 04:41 |
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Function names are case-insensitive in PHP 3 and PHP 4.
For
example, if you have function foo, you can call it using foo(), FoO(),
FOO(), etc..
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[email protected]
09-Mar-2001 04:42 |
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PHP supports recursion. I thought it worth to mention.
Simple
Quick Sort using recursion works perfectly.
== OUTPUT
==
Recursion TEST
Array
(
[0] => 12
[1] => 23
[2] => 35
[3] => 45
[4] =>
56
[5] => 67
)
== END OUTPUT ==
== QUICK SORT
CODE ==
<?php
echo('
Recursion
TEST
');
function swap(&$v, $i, $j) {
$temp =
$v[$i];
$v[$i] = $v[$j];
$v[$j] = $temp;
}
// Quick
Sort integer array - $int_array[$left] .. $int_array[$right]
function
qsort(&$int_array, $left, $right) {
if ($left >= $right)
return; // Do nothing if there are less than 2 array elements
swap
($int_array, $left, intval(($left+$right)/2));
$last = $left;
for ($i = $left + 1; $i <= $right; $i++)
if ($int_array[$i] <
$int_array[$left])
swap($int_array, ++$last, $i);
swap($int_array, $left, $last);
qsort($int_array, $left,
$last-1);
qsort($int_array, $last+1, $right);
}
$val =
array(56,23,45,67,12,35);
qsort($val, 0,
count($val)-1);
echo '<pre>';
print_r ($val);
echo
'</pre>';
?>
== END QUICK SORT ==
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[email protected]
05-Apr-2001 02:34 |
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[Editor's note: put your includes in the beginning of your script. You can
call an included function, after it has been included
--jeroen]
The documentation states: "In PHP 3, functions
must be defined before they are referenced. No such requirement exists in
PHP 4."
I thought it wise to note here that there is in fact
a limitation: you cannot bury your function in an include() or require().
If the function is in an include()'d file, there is NO way to call that
function beforehand. The workaround is to put the function directly in
the file that calls the function.
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[email protected]
14-Nov-2001 06:47 |
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It is possible to define a function from inside another function.
The
result is that the inner function does not exist until the outer function
gets executed.
For example, the following
code:
function a () {
function b() {
echo "I
am b.\n";
}
echo "I am a.\n";
}
if
(function_exists("b")) echo "b is defined.\n"; else
echo "b is not defined.\n";
a();
if
(function_exists("b")) echo "b is defined.\n"; else
echo "b is not defined.\n";
echoes:
b is not
defined.
I am a.
b is defined.
Classes too can be defined
inside functions, and will not exist until the outer function gets
executed.
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[email protected]
14-Feb-2002 12:57 |
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As a corollary to other people's contributions in this section, you have to
be careful when transforming a piece of code in to a function (say F1). If
this piece of code contains calls to another function (say F2), then each
variable used in F2 and defined in F1 must be declared as GLOBAL both in
F1 and F2. This is tricky.
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bishop
01-May-2002 01:49 |
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The documentation statement:
"In PHP 3, functions must be
defined before they are referenced. No such requirement exists in PHP
4."
is not entirely accurate (or at least
misleading).
Consider:
function a() {
function b() {
echo 'I am b';
}
echo 'I am a';
}
You MUST
call a() before you can even think about using b(). Why? The parser
hasn't touched the scope inside function a (for efficiency reasons), so b
has yet to be defined or even *declared*.
The documentation (I
believe) refers to this behaviour:
a();
function a() {
echo
'I am a';
}
which is perfectly valid and runs as you probably
expect. However, the following does not work as expected (or
documented):
function a() {
b();
function b()
{
echo 'I am b';
}
echo 'I am
a';
}
a();
Rather than getting "I am b"
followed by "I am a" you get a parse error ("Call to
undefined function b()").
So, the bottom line:
Gewaehrleistungsausschluss
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bishop
01-May-2002 01:54 |
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Consider:
function a() {
function b() {
echo 'I am
b';
}
echo 'I am a';
}
a();
a();
As you
might NOT expect, the second call to a() fails with a "Cannot
redeclare b()" error. This behaviour is correct, insofar as PHP
doesn't "allow functions to be redefined."
A work
around:
function a() {
if ( ! function_exists('b') ) {
function b() {
echo 'I am b';
}
}
echo 'I am a';
}
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[email protected]
20-Jun-2002 03:51 |
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I have some problems with functions. I want to create some kind of 'tell me
the configuration function'. Like this:
<?php
function
TellConfig(){
$header_file_to_use =
"header.inc";
$footer_file_to_user =
"footer.inc";
}
//Wont
Work!
TellConfig();
include($header_file_to_use);
echo
"Body!";
include($footer_file_to_use);
/*
Is there a
way to call a variable set by a function? As i dont know this is the way i
am doing it
*/
$header_file_to_use =
"";
$footer_file_to_use = "";
function
TellConfig2(){
global $header_file_to_use,
$footer_file_to_use;
$header_file_to_use =
"header.inc";
$footer_file_to_use =
"footer.inc";
}
TellConfig2();
include($header_file_to_use);
echo
"body 2!";
include($footer_file_to_use);
// This
works
?>
Even i dont want to use a function for configuration
vars this is a good example.
:P
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[email protected]
10-Aug-2002 01:17 |
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Although [email protected] has already pointed the recursion support
on PHP, here's another example, wich shows clearly the mechanism of
recursive algorithms:
function fact($i){
if($i==1){
return 1;
}else{
return $i*fact($i-1);
}
}
It
returns $i! (supposing $i is a valid positive integer greater than 0).
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[email protected]
13-Aug-2002 11:26 |
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[Editor's note: that is the reason why 'function' and others are
"reserved words", see:
for more details]
Creating a function with the name 'function'
won't turn up errors, but when calling that function, nothing will be
returned. Example:
<?php
function();
function
function () {
print "Hello, world!";
}
//
"Hello, world!" will not be printed.
?>
I assume
this is because when calling 'function', php thinks you are trying to
create a new function.
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include_once