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位运算符> <算术运算符
[edit] Last updated: Fri, 25 May 2012

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赋值运算符

基本的赋值运算符是“=”。一开始可能会以为它是“等于”,其实不是的。它实际上意味着把右边表达式的值赋给左边的运算数。

赋值运算表达式的值也就是所赋的值。也就是说,“$a = 3”的值是 3。这样就可以做一些小技巧:

<?php

$a 
= ($b 4) + 5// $a 现在成了 9,而 $b 成了 4。

?>

在基本赋值运算符之外,还有适合于所有二元算术,数组集合和字符串运算符的“组合运算符”,这样可以在一个表达式中使用它的值并把表达式的结果赋给它,例如:

<?php

$a 
3;
$a += 5// sets $a to 8, as if we had said: $a = $a + 5;
$b "Hello ";
$b .= "There!"// sets $b to "Hello There!", just like $b = $b . "There!";

?>

注意赋值运算将原变量的值拷贝到新变量中(传值赋值),所以改变其中一个并不影响另一个。这也适合于在很密集的循环中拷贝一些值例如大数组。也可以使用引用赋值,用 $var = &$othervar; 语法。引用赋值意味着两个变量都指向同一个数据,没有任何数据的拷贝。有关引用的更多信息见引用的说明。在 PHP 5中,对象总是通过引用赋值的,除非明确使用新的 clone关键字。



位运算符> <算术运算符
[edit] Last updated: Fri, 25 May 2012
 
add a note add a note User Contributed Notes 赋值运算符
haubertj at alfredstate dot edu 01-Sep-2011 07:19
[[   Editor's note: You are much better off using the foreach (array_expression as $key => $value) control structure in this case   ]]

When using

<php
while ($var = current($array) {
#do stuff
next($aray)
?>

to process an array, if current($array) happens to be falsy but not === false it will still end the loop.  In such a case strict typing must be used.

Like this:

<php
while (($var = current($array)) !== FALSE) {
#do stuff
next($aray)
?>

Of course if your array may contain actual FALSE values you will have to deal with those some other way.
Peter, Moscow 11-Feb-2011 01:44
Using $text .= "additional text"; instead of $text =  $text ."additional text"; can seriously enhance performance due to memory allocation efficiency.

I reduced execution time from 5 sec to .5 sec (10 times) by simply switching to the first pattern for a loop with 900 iterations over a string $text that reaches 800K by the end.
Hayley Watson 05-Feb-2008 05:54
You could also take adam at gmail dot com's xor-assignment operator and use the fact that it's right-associative:

$a ^= $b ^= $a ^= $b;
Hayley Watson 07-Oct-2007 03:22
bradlis7 at bradlis7 dot com's description is a bit confusing. Here it is rephrased.

<?php
$a
= 'a';
$b = 'b';

$a .= $b .= "foo";

echo
$a,"\n",$b;?>
outputs

abfoo
bfoo

Because the assignment operators are right-associative and evaluate to the result of the assignment
<?php
$a
.= $b .= "foo";
?>
is equivalent to
<?php
$a
.= ($b .= "foo");
?>
and therefore
<?php
$b
.= "foo";
$a .= $b;
?>
adam at gmail dot com 25-Aug-2006 10:38
or you could use the xor-assignment operator..
$a ^= $b;
$b ^= $a;
$a ^= $b;
bradlis7 at bradlis7 dot com 15-Aug-2005 08:13
Note whenever you do this

<?php
$a
.= $b .= "bla bla";
?>

it comes out to be the same as the following:

<?php
$a
.= $b."bla bla";
$b .= "bla bla";
?>

So $a actually becomes $a and the final $b string. I'm sure it's the same with numerical assignments (+=, *=...).
straz at mac dot nospam dot com 20-Feb-2004 10:18
This page really ought to have table of assignment operators,
namely,

See the Arithmetic Operators page (http://www.php.net/manual/en/language.operators.arithmetic.php)
Assignment    Same as:
$a += $b     $a = $a + $b    Addition
$a -= $b     $a = $a - $b     Subtraction
$a *= $b     $a = $a * $b     Multiplication
$a /= $b     $a = $a / $b    Division
$a %= $b     $a = $a % $b    Modulus

See the String Operators page(http://www.php.net/manual/en/language.operators.string.php)
$a .= $b     $a = $a . $b       Concatenate

See the Bitwise Operators page (http://www.php.net/manual/en/language.operators.bitwise.php)
$a &= $b     $a = $a & $b     Bitwise And
$a |= $b     $a = $a | $b      Bitwise Or
$a ^= $b     $a = $a ^ $b       Bitwise Xor
$a <<= $b     $a = $a << $b     Left shift
$a >>= $b     $a = $a >> $b      Right shift

 
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