PHP: Inkrement- bzw. Dekrementoperatoren

Last updated: Fri, 30 Oct 2009

Inkrement- bzw. Dekrementoperatoren

PHP unterstützt Prä- und Post-Inkrement- und Dekrementoperatoren im Stil der Programmiersprache C.

**Inkrement- und Dekrementoperatoren**
Beispiel	Name	Auswirkung
++$a	Prä-Inkrement	Erhöht den Wert von $a um eins (inkrementiert $a) und gibt anschließend den neuen Wert von $a zurück.
$a++	Post-Inkrement	Gibt zuerst den aktuellen Wert von $a zurück und erhöht dann den Wert von $a um eins.
--$a	Prä-Dekrement	Vermindert den Wert von $a um eins (dekrementiert $a) und gibt anschließend den neuen Wert von $a zurück.
$a--	Post-Dekrement	Gibt zuerst den aktuellen Wert von $a zurück und erniedrigt dann den Wert von $a um eins.

Ein einfaches Beispiel-Skript:


<?php
echo "<h3>Post-Inkrement</h3>";
$a = 5;
echo "Sollte 5 sein: " . $a++ . "<br />\n";
echo "Sollte 6 sein: " . $a . "<br />\n";

echo "<h3>Pre-Inkrement</h3>";
$a = 5;
echo "Sollte 6 sein: " . ++$a . "<br />\n";
echo "Sollte 6 sein: " . $a . "<br />\n";

echo "<h3>Post-Dekrement</h3>";
$a = 5;
echo "Sollte 5 sein: " . $a-- . "<br />\n";
echo "Sollte 4 sein: " . $a . "<br />\n";

echo "<h3>Pre-Dekrement</h3>";
$a = 5;
echo "Sollte 4 sein: " . --$a . "<br />\n";
echo "Sollte 4 sein: " . $a . "<br />\n";
?>

PHP folgt bei der Behandlung arithmetischer Operationen an Zeichenvariablen der Perl-Konvention und nicht der von C. Zum Beispiel wird in Perl aus 'Z'+1 'AA', während aus 'Z'+1 in C '[' wird ( ord('Z') == 90, ord('[') == 91 ). Beachten Sie, dass Zeichenvariablen zwar inkrementiert aber nicht dekrementiert werden können.

Beispiel #1 Arithmetrische Operationen an Zeichenvariablen


<?php
$i = 'W';
for($n=0; $n<6; $n++)
  echo ++$i . "\n";

/*
  Erzeugt in etwa folgende Ausgabe:

X
Y
Z
AA
AB
AC

*/
?>

Logische Operatoren

Operatoren zur Programmausführung

Last updated: Fri, 30 Oct 2009

add a note User Contributed Notes
Inkrement- bzw. Dekrementoperatoren

daevid at daevid dot com
02-Oct-2009 09:11


<?php

$num = 123;

$num = $num++;

print $num;  //this prints 123 and not 124 ?!!



$num = 123;

$num = ++$num;

print $num; //this prints 124 as expected



$num = 123;

$num++;

print $num;  //this prints 124 as expected



$num = 123;

print $num++; //this prints 123 and not 124 ?!!

print $num++; //this NOW prints 124

?>



So then I read the manual because I think I'm loosing my mind and perhaps it's backwards day and nobody told me.



I'm baffled as to the reasoning behind:

"$a++  :: Post-increment :: Returns $a, then increments $a by one."



Why would you EVER want $num = $num++; to give you back the value you already had? Even if we did $foo = $bar++; I would still logically (and common sensely) expect $foo to be the increment of $bar!



It also seems counter-intuitive, as I was always lead to believe everything is processed on the right of an equals sign and then assigned back to the left side of the equals sign. In this case, it does a mixture of both.

dsbeam at gmail dot com
31-Aug-2009 10:35


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 10:49


(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 08:15


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 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


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


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


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


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


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


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


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


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


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


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"

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Logische Operatoren

Operatoren zur Programmausführung

Last updated: Fri, 30 Oct 2009