执行运算符
Note that incrementing or decrementing arrays or objects variables have no effects.
<?php
$a = array(0,'1','foo'=>'bar',3.5);
$a++;
print_r($a);
?>
gives the same result that
<?php
$a = array(0,'1','foo'=>'bar',3.5);
print_r($a);
?>
gives.
Logical, but, need to be noticed.
递增／递减运算符
PHP 支持 C 风格的前／后递增与递减运算符。
Note: 递增／递减运算符不影响布尔值。递减
NULL值也没有效果，但是递增NULL的结果是 1。
| 例子 | 名称 | 效果 |
|---|---|---|
| ++$a | 前加 | $a 的值加一，然后返回 $a。 |
| $a++ | 后加 | 返回 $a，然后将 $a 的值加一。 |
| --$a | 前减 | $a 的值减一， 然后返回 $a。 |
| $a-- | 后减 | 返回 $a，然后将 $a 的值减一。 |
一个简单的示例脚本：
<?php
echo "<h3>Postincrement</h3>";
$a = 5;
echo "Should be 5: " . $a++ . "<br />\n";
echo "Should be 6: " . $a . "<br />\n";
echo "<h3>Preincrement</h3>";
$a = 5;
echo "Should be 6: " . ++$a . "<br />\n";
echo "Should be 6: " . $a . "<br />\n";
echo "<h3>Postdecrement</h3>";
$a = 5;
echo "Should be 5: " . $a-- . "<br />\n";
echo "Should be 4: " . $a . "<br />\n";
echo "<h3>Predecrement</h3>";
$a = 5;
echo "Should be 4: " . --$a . "<br />\n";
echo "Should be 4: " . $a . "<br />\n";
?>
在处理字符变量的算数运算时，PHP 沿袭了 Perl 的习惯，而非 C 的。例如，在 Perl 中 'Z'+1 将得到 'AA'，而在 C 中，'Z'+1 将得到 '['（ord('Z') == 90，ord('[') == 91）。注意字符变量只能递增，不能递减，并且只支持纯字母（a-z 和 A-Z）。
Example #1 涉及字符变量的算数运算
<?php
$i = 'W';
for ($n=0; $n<6; $n++) {
echo ++$i . "\n";
}
?>
以上例程会输出：
X Y Z AA AB AC
递增或递减布尔值没有效果。
add a note
User Contributed Notes
递增／递减运算符
nicolas dot giraud at actiane dot com
20-Apr-2012 09:12
johnnythan at nospam dot gmx dot com
22-Nov-2011 04:05
If you have a trailing zero and use the increment, the trailing zero will not remain. Was at least unexpected for me at first, although it's logical if you think about it.
<?php
$start = '01';
$start++;
print $start; //Outputs '2' not '02'
?>
Anonymous
03-Oct-2011 02:16
Notice the different output of the following two for-loops.
<?php
for ($i='a'; $i<='y'; $i++) {
echo "$i ";
}
?>
This one will output the characters 'a' to 'y'.
<?php
for ($i='a'; $i<='z'; $i++) {
echo "$i ";
}
?>
This second one will output the characters 'a' to 'z' AND after that the strings 'aa', ..., 'az', 'ba', ..., 'bz', 'ca', ... and so on to 'yz'.
Brad Proctor
06-Nov-2010 08:51
I ran some tests (on PHP 5.3.3) of my own and was surprised to find $i += 1 to be the fastest method of incrementing. Here are the methods fastest to slowest:
$i += 1;
++$i;
$i++;
$i = $i + 1;
dsbeam at gmail dot com
31-Aug-2009 03:35
When using the ++ operator by itself on a variable, ++$var is faster than $var++ and uses slightly less memory (in my experiments). It would seem like this could be optimized in the language during runtime (if $var++ is the only thing in the whole statement, it could be treated as ++$var).
I conducted many tests (I believe to be fair), and here's one of the results:
$i++ took 8.47515535355 seconds and 2360 bytes
++$i took 7.80081486702 seconds and 2160 bytes
Here's my code. If anyone sees a bias in it, tell me. I conducted it many times, each time going through a loop one million iterations and doing each test 10 - 15 times (10 - 15 million uses of the ++ operator).
<?php
ini_set( 'MAX_EXEC_TIME', 120 );
ob_start( );
$num_tests = 10;
$startFirst = $startSecond = $endFirst = $endSecond = $startFirstMemory = $endFirstMemory = $startSecondMemory = $endSecondMemory = $someVal = 0;
$times = array( '$i++' => array( 'time' => 0, 'memory' => 0 ), '++$i' => array( 'total' => 0, 'memory' => 0 ) );
for( $j = 0; $j < $num_tests; ++$j )
{
for( $i = 0, $startFirstMemory = memory_get_usage( ), $startFirst = microtime( true ); $i < 10000000; $i++ ){ $someval = 2; }
$endFirstMemory = memory_get_usage( );
$endFirst = microtime( true );
for( $i = 0, $startSecondMemory = memory_get_usage( ), $startSecond = microtime( true ); $i < 10000000; ++$i ){ $someval = 2; }
$endSecondMemory = memory_get_usage( );
$endSecond = microtime( true );
$times[ '$i++' ][ $j ] = array( 'startTime' => $startFirst, 'endTime' => $endFirst, 'startMemory' => $startFirstMemory, 'endMemory' => $endFirstMemory );
$times[ '++$i' ][ $j ] = array( 'startTime' => $startSecond, 'endTime' => $endSecond, 'startMemory' => $startSecondMemory, 'endMemory' => $endSecondMemory );
}
for( $i = 0; $i < $num_tests; ++$i )
{
$times[ '$i++' ][ 'time' ] += ( $times[ '$i++' ][ $i ][ 'endTime' ] - $times[ '$i++' ][ $i ][ 'startTime' ] );
$times[ '++$i' ][ 'time' ] += ( $times[ '++$i' ][ $i ][ 'endTime' ] - $times[ '++$i' ][ $i ][ 'startTime' ] );
$times[ '$i++' ][ 'memory' ] += ( $times[ '$i++' ][ $i ][ 'endMemory' ] - $times[ '$i++' ][ $i ][ 'startMemory' ] );
$times[ '++$i' ][ 'memory' ] += ( $times[ '++$i' ][ $i ][ 'endMemory' ] - $times[ '++$i' ][ $i ][ 'startMemory' ] );
}
echo 'There were ' . $num_tests . ' tests conducted, here\'s the totals<br /><br />
$i++ took ' . $times[ '$i++' ][ 'time' ] . ' seconds and ' . $times[ '$i++' ][ 'memory' ] . ' bytes<br />
++$i took ' . $times[ '++$i' ][ 'time' ] . ' seconds and ' . $times[ '++$i' ][ 'memory' ] . ' bytes';
ob_end_flush( );
?>
Try it yourself, ;)
sneskid at hotmail dot com
07-Aug-2009 03:49
(related to what "Are Pedersen" wrote)
With arrays it can lead to much confusion if your index variable is altered on the right side of the = sign, either with ++|-- or even when passed to a function by reference..
Consider these (PHP 5):
<?php
$A[$a] = ++$a; // [1]=1
$B[++$b] = ++$b; // [1]=2
$C[$c+=0] = ++$c; // [0]=1
?>
In 'A' you have to be aware that PHP evaluates $A[$a] last.
Yet in 'B' and 'C' PHP evaluates the index and saves it in a temporary variable.
You can always force PHP to evaluate a variable without explicitly storing it as a named variable first, with a simple "+=0" like in example 'C'.
Compared to 'A', 'C' gives the more logically expected result, when we expect evaluation occurs left to right.
PHP does evaluate left to right BUT it will attempt to cut down on temporary variables, which can lead to confusing results.
So just be aware and use either behavior to your advantage for the desired functionality.
pov at fingerprint dot fr
28-Mar-2008 12:15
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 $var. So PHP5 got right, and it's not a "strange behaviour", it's only a solved bug.
Anyway, it's always a bad idea to pass anything other than a variable as a by-reference parameter...
Anonymous
09-Jan-2008 03:17
Some strange behaviour between PHP 4 and 5.
Code :
<?php
function change (&$var) {
$var += 10;
}
$var = 1;
++$var;
change($var);
echo "var=$var";
$var = 1;
change(++$var);
echo "var=$var";
?>
Output in PHP4
var=12
var=12
Output in PHP5
var=12
var=2
michal dot kocarek at NO_SPAM dot seznam dot cz
23-Sep-2007 03:04
Speed tip:
Do not use post-incrementation/post-decrementation ($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 07:34
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 06:52
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 03:08
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 = $a + 2 + 2
Answer: 3 + 2 + 2 = 7
julien-bernie-laurent at polenord.com
01-Mar-2006 07:55
to thus trying to increment a string and are blocked by the exponential typecast explained in the message below, here is a small function :
function increment($var) {
$var2 = '_'.$var;
return substr(++$var2,1);
}
timo at frenay dot net
25-Aug-2004 08:45
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 01:19
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 04:11
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 09:11
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 + $i++;
Result: 4. The first i is 2. Second i is 2 too, gets incremented afterwards.
cleong at letstalk dot com
17-Oct-2001 07:52
Note that the ++ and -- don't convert a boolean to an int. The following code will loop forever.
function a($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 12:02
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"