実行演算子
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 $var. So PHP5 got right, and it's not a "strange behaviour", it's only a solved bug.
Anyway, it's always a bad idea to pass anything other than a variable as a by-reference parameter...
加算子/減算子
PHP は C 言語形式の加算子/減算子（前置・後置ともに）をサポートします。
注意: 加算子/減算子は bool 型の値には何も変更を加えません。 同じく NULL に減算子を適用しても何も起こりませんが、NULL に加算子を 適用すると 1 となります。
| 例 | 名前 | 効果 |
|---|---|---|
| ++$a | 前置加算子 | $a に 1 を加え、$a を返します。 |
| $a++ | 後置加算子 | $a を返し、$a に1を加えます。 |
| --$a | 前置減算子 | $a から 1 を引き、$a を返します。 |
| $a-- | 後置減算子 | $aを返し、$a から 1 を引きます。 |
以下に簡単なスクリプトの例を示します。
<?php
echo "<h3>後置加算</h3>";
$a = 5;
echo "5 となります: " . $a++ . "<br>\n";
echo "6 となります: " . $a . "<br>\n";
echo "<h3>前置加算</h3>";
$a = 5;
echo "6 となります: " . ++$a . "<br>\n";
echo "6 となります: " . $a . "<br>\n";
echo "<h3>後置減算</h3>";
$a = 5;
echo "5 となります: " . $a-- . "<br>\n";
echo "4 となります: " . $a . "<br>\n";
echo "<h3>前置減算</h3>";
$a = 5;
echo "4 となります: " . --$a . "<br>\n";
echo "4 となります: " . $a . "<br>\n";
?>
PHP は、算術演算子で文字変数を扱った場合に C ではなく Perl の慣習に 従います。例えば、perl では 'Z'+1 は 'AA' を返しますが C では 'Z'+1 は '[' ( ord('Z') == 90, ord('[') == 91 ) を返します。 文字変数はインクリメントされることは可能ですがデクリメントは不可能であるということ、 またプレーンな ASCII 文字 (a-z および A-Z) のみがサポートされるということに注意しましょう。
例1 文字変数に対する算術演算子の使用
<?php
$i = 'W';
for ($n=0; $n<6; $n++) {
echo ++$i . "\n";
}
?>
上の例の出力は以下となります。
X Y Z AA AB AC
論理型に対する加算/減算は何の影響も及ぼしません。
add a note
User Contributed Notes加算子/減算子
pov at fingerprint dot fr
28-Mar-2008 08:15
28-Mar-2008 08:15
Anonymous
09-Jan-2008 11:17
09-Jan-2008 11:17
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=$var";
?>
Output in PHP4
var=12
var=12
Output in PHP5
var=12
var=2
michal dot kocarek at NO_SPAM dot seznam dot cz
23-Sep-2007 10:04
23-Sep-2007 10:04
Speed tip:
Do not use post-incrementation/post-decrementation ($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 02:34
14-Jun-2007 02:34
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
21-Apr-2007 01:52
21-Apr-2007 01:52
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 11:08
28-Feb-2007 11:08
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 = $a + 2 + 2
Answer: 3 + 2 + 2 = 7
julien-bernie-laurent at polenord.com
01-Mar-2006 03:55
01-Mar-2006 03:55
to thus trying to increment a string and are blocked by the exponential typecast explained in the message below, here is a small function :
function increment($var) {
$var2 = '_'.$var;
return substr(++$var2,1);
}
timo at frenay dot net
25-Aug-2004 03:45
25-Aug-2004 03:45
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 09:19
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
07-Feb-2004 12:11
07-Feb-2004 12:11
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
15-Oct-2002 04:11
15-Oct-2002 04:11
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 + $i++;
Result: 4. The first i is 2. Second i is 2 too, gets incremented afterwards.
cleong at letstalk dot com
18-Oct-2001 02:52
18-Oct-2001 02:52
Note that the ++ and -- don't convert a boolean to an int. The following code will loop forever.
function a($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 07:02
18-Jul-2001 07:02
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"