Operatori di esecuzione
In reply to Anonymous :
What is strange is that you didn't get an error : ++$var is an expression and can't therefore not be referenced.
Now, if you suppose an implicit assignment to an invisible variable, your code becomes :
<?php
$var = 1;
$plus_plus_var = ++$var;
change($plus_plus_var);
echo "var=$var";
?>
Written as such, change clearly acts on $plus_plus_var, not on $var. So PHP5 got right, and it's not a "strange behaviour", it's only a solved bug.
Anyway, it's always a bad idea to pass anything other than a variable as a by-reference parameter...
Operatori di incremento/decremento
PHP supporta lo stile C degli operatori di pre- e post-incremento e decremento.
Nota: Gli operatori di incremento e decremento non agiscono sui valori boolean. Anche decrementare il valore NULL non ha effetti, ma incrementarlo darà come risultato 1.
| Esempio | Nome | Effetto |
|---|---|---|
| ++$a | Pre-incremento | Incrementa $a di una unità, inoltre restituisce $a. |
| $a++ | Post-incremento | Restituisce $a, inoltre incrementa $a di una unità. |
| --$a | Pre-decremento | Decrementa $a di una unità, inoltre restituisce $a. |
| $a-- | Post-decremento | Restituisce $a, inoltre decrementa $a di una unità. |
Qui c'è un semplice script di esempio:
<?php
echo "<h3>Postincrement</h3>";
$a = 5;
echo "Dovrebbe essere 5: " . $a++ . "<br />\n";
echo "Dovrebbe essere 6: " . $a . "<br />\n";
echo "<h3>Pre-incremento</h3>";
$a = 5;
echo "Dovrebbe essere 6: " . ++$a . "<br />\n";
echo "Dovrebbe essere 6: " . $a . "<br />\n";
echo "<h3>Post-decremento</h3>";
$a = 5;
echo "Dovrebbe essere 5: " . $a-- . "<br />\n";
echo "Dovrebbe essere 4: " . $a . "<br />\n";
echo "<h3>Pre-decremento</h3>";
$a = 5;
echo "Dovrebbe essere 4: " . --$a . "<br />\n";
echo "Dovrebbe essere 4: " . $a . "<br />\n";
?>
Il PHP segue le convenzioni di Perl e non del C quando tratta le operazioni matematiche sui caratteri. Ad esempio, in Perl 'Z'+1 restituisce 'AA', mentre in C 'Z'+1 restituisce '[' ( ord('Z') == 90, ord('[') == 91 ). Attenzione che le variabili con caratteri possono essere sommate e non sottratte.
Example #1 Operazioni aritmetiche su variabili stringa
<?php
$i = 'W';
for ($n=0; $n<6; $n++)
echo ++$i . "\n";
Il precedente esempio visualizzerà:
X Y Z AA AB AC
Incrementare o decrementare valori booleani non ha effetto.
add a note
User Contributed NotesOperatori di incremento/decremento
pov at fingerprint dot fr
28-Mar-2008 02:15
28-Mar-2008 02:15
Anonymous
09-Jan-2008 04:17
09-Jan-2008 04:17
Some strange behaviour between PHP 4 and 5.
Code :
<?php
function change (&$var) {
$var += 10;
}
$var = 1;
++$var;
change($var);
echo "var=$var";
$var = 1;
change(++$var);
echo "var=$var";
?>
Output in PHP4
var=12
var=12
Output in PHP5
var=12
var=2
michal dot kocarek at NO_SPAM dot seznam dot cz
23-Sep-2007 04:04
23-Sep-2007 04:04
Speed tip:
Do not use post-incrementation/post-decrementation ($i++, $i--) where you do not work with the result of this expression.
(For novices: Yes, every expression returns an result, also $a = '5' returns result, same as $a && $b. And this consumes more time and resources.)
When writing loops, replace the post-incrementation with pre-incrementation, it is around 3times faster than post-incrementation.
Why? In post-incrementation, PHP needs to copy variable value somewhere, then it increments the value, then returns the value which was stored before the incrementation was done. No matter if you don't expect the return value, PHP is scripting language, not compiled one, so it doesn't optimize use of return values.
<?php
// Good practice for loop:
$array_count = count($array); // Store temporarily instead of calling everytime in loop
for ($i = 0; $i < $max_count; ++$i) { // Use pre-incrementation here, it is faster
// do something here
}
?>
rowan dot collins at gmail dot com
14-Jun-2007 08:34
14-Jun-2007 08:34
As the manual says, decrementing NULL in this way yields NULL, although incrementing it yields 1, as you might expect. Can't quite see why this makes sense, but if you need to work around it, you can use '-= 1' instead:
<?php
$i = null;
--$i;
var_dump($i); // NULL
$i--;
var_dump($i); // NULL
$i-=1;
var_dump($i); // int(-1)
?>
Note that -= returns the value assigned, so treat it like '--$i', not '$i--' if you're testing the value.
Q1712 at online dot ms
20-Apr-2007 07:52
20-Apr-2007 07:52
A more detailed explanation of the string incremant is:
First of all it is checked wether the string is a standart representaion of a number wich is true if it equals the regex /^ *[+-]?[0-9]*(\.[0-9]|[0-9]\.)[0-9]*([eE]?[+-]?[0-9]+)?$/
but not the regex /\+\./ (no idea why).
if it does, the type is changed to integer (if it equals /^ *[+-]?[0-9]+$/) or to float and then incremented by one.
An empty string becomes the string "1".
Otherwise if the last character is one of [0-8], [a-y] or [A-Y] it is incremented. If it is Z it puts it back to A, is z to a, if 9 to 0 and trys to do the same with the previouse character.
If a character is reatched that is not in [0-9a-zA-Z], nothing is done anymore (that's why " Z" will increment to " A").
If the begining is reached a new caracter is prepended. "1" "a" or "A" depending on wether the first character was "9", "z" or "Z".
If the last character was not [0-9a-zA-Z] the string isn't chaged.
hope this helps someone
Are Pedersen
28-Feb-2007 04:08
28-Feb-2007 04:08
Something to think about:
$a=1;
$a += $a++ + ++$a;
echo $a;
will give you 7.
Why is this?
1. ++$a is first incremented. Now $a is 2.
$a += $a++ + 2
$a is 2
2. $a++ is added to 2 then $a is incremented
$a += 2 + 2
$a is 3
3. now the value of 2 + 2 is added to $a ($a is 3)
$a = $a + 2 + 2
Answer: 3 + 2 + 2 = 7
julien-bernie-laurent at polenord.com
01-Mar-2006 08:55
01-Mar-2006 08:55
to thus trying to increment a string and are blocked by the exponential typecast explained in the message below, here is a small function :
function increment($var) {
$var2 = '_'.$var;
return substr(++$var2,1);
}
timo at frenay dot net
25-Aug-2004 09:45
25-Aug-2004 09:45
JMcCarthy AT CitiStreet DOT com:
As for your March 31 post, at least in PHP version 4.3 this no longer holds for 'D'. Your point is still valid for 'e' or 'E' and worth noting.
Your comment from May 12 is simply not true, although it might be a bug in your specific version of PHP but that would seem very strange.
<?php
$Align = array('a', 'b', 'c');
$i = 0;
echo $Align[$i++]; // Prints 'a', as expected
?>
It might be interesting to know that pre-/postincrement assumes a value of 0 for undefined variables, but pre-/postdecrement does not:
<?php
echo var_dump(++$foo); // int(1)
echo var_dump(--$bar); // NULL!
?>
31-Mar-2004 02:19
Note that incrementing strings can give unpredictable results due to type changes. For example:
<?php
$i = '9C6';
for($n=0; $n<10; $n++)
echo ++$i . "\n";
?>
Gives you:
9C7
9C8
9C9
9D0
10
11
12
..etc.
The 'D' (and also 'E') characters are interpreted here as exponents of 10 (i.e., scientific notation) formatted numbers. Using '9D6' will give 9000001, 9000002, etc.
You might want to use all alphabetical or all numerical, but not mix the two otherwise you may not get what you expect..
chris at free-source dot com
06-Feb-2004 05:11
06-Feb-2004 05:11
Interesting performance note:
$i++ seems to be slightly slower than ++$i, when used on a line by itself the 2 have the same purpose. It's not much, but over 100,000 incements the pre-increment is about .004 seconds faster on average.
mu at despammed dot net
14-Oct-2002 10:11
14-Oct-2002 10:11
The exact moment when post-increment and post-decrement happen is _just immediately after the variable is evaluated_ (not "after the line is processed" or something like that)
Example 1:
$i = 2;
echo $i++ + $i;
Result: 5. The first i is evaluated as 2, gets incremented to 3. i is then evaluated as 3 for the second occurance.
Example 2:
$i = 2;
echo $i + $i++;
Result: 4. The first i is 2. Second i is 2 too, gets incremented afterwards.
cleong at letstalk dot com
17-Oct-2001 08:52
17-Oct-2001 08:52
Note that the ++ and -- don't convert a boolean to an int. The following code will loop forever.
function a($start_index) {
for($i = $start_index; $i < 10; $i++) echo "\$i = $i\n";
}
a(false);
This behavior is, of course, very different from that in C. Had me pulling out my hair for a while.
fred at surleau dot com
18-Jul-2001 01:02
18-Jul-2001 01:02
Other samples :
$l="A"; $l++; -> $l="B"
$l="A0"; $l++; -> $l="A1"
$l="A9"; $l++; -> $l="B0"
$l="Z99"; $l++; -> $l="AA00"
$l="5Z9"; $l++; -> $l="6A0"
$l="9Z9"; $l++; -> $l="10A0"
$l="9z9"; $l++; -> $l="10a0"
$l="J85410"; $l++; -> $l="J85411"
$l="J99999"; $l++; -> $l="K00000"
$l="K00000"; $l++; -> $l="K00001"