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运算符是可以通过给出的一或多个值(用编程行话来说,表达式)来产生另一个值(因而整个结构成为一个表达式)的东西。
运算符可按照其能接受几个值来分组。一元运算符只能接受一个值,例如 !(逻辑取反运算符)或 ++(递增运算符)。二元运算符可接受两个值,例如熟悉的算术运算符 +(加)和 -(减),大多数 PHP 运算符都是这种。最后是唯一的三元运算符 ? :,可接受三个值;通常就简单称之为“三元运算符”(尽管称之为条件运算符可能更合适)。
PHP 的运算符完整列表见下节运算符优先级。该节也解释了运算符优先级和结合方向,这控制着在表达式包含有若干个不同运算符时究竟怎样对其求值。
[#1] arth dot inbox+php dot net at gmail dot com [2015-06-12 14:51:07]
&= operator.
It is binary and followed expressions give different results:
$a = $b = 1;
echo $a &= 2; // 0
echo $a = $a && 2; // 1
[#2] pgarvin76+php dot net at NOSPAMgmail dot com [2008-12-29 16:43:52]
Method chaining is read left to right (left associative):
<?php
class Test_Method_Chain
{
public function One()
{
echo "One" . PHP_EOL;
return $this;
}
public function Two()
{
echo "Two" . PHP_EOL;
return $this;
}
public function Three()
{
echo "Three" . PHP_EOL;
return $this;
}
}
$test = new Test_Method_Chain();
$test->One()->Two()->Three();
?>
[#3] ddascalescu at gmail dot com [2008-10-23 18:53:20]
The -> operator, not listed above, is called "object operator" (T_OBJECT_OPERATOR).
[#4] figroc at gmail dot com [2008-08-02 03:30:22]
The variable symbol '$' should be considered as the highest-precedence operator, so that the variable variables such as $$a[0] won't confuse the parser. [http://www.php.net/manual/en/language.variables.variable.php]
[#5] phpnet dot 20 dot dpnsubs at xoxy dot net [2007-11-01 14:13:30]
Note that in php the ternary operator ?: has a left associativity unlike in C and C++ where it has right associativity.
You cannot write code like this (as you may have accustomed to in C/C++):
<?php
$a = 2;
echo (
$a == 1 ? 'one' :
$a == 2 ? 'two' :
$a == 3 ? 'three' :
$a == 4 ? 'four' : 'other');
echo "\n";
// prints 'four'
?>
You need to add brackets to get the results you want:
<?php
$a = 2;
echo ($a == 1 ? 'one' :
($a == 2 ? 'two' :
($a == 3 ? 'three' :
($a == 4 ? 'four' : 'other') ) ) );
echo "\n";
//prints 'two'
?>
[#6] Gautam [2007-10-10 03:22:19]
<?php
$result1 = 7 + 8 * 9/3 -4;
$result2 = 7 + 8 * (9/3 -4);
$result3 =(7 + 8)* 9/3 -4;
echo "Result1 for 7 + 8 * 9/3 -4 = $result1 Result2 for 7 + 8 * (9/3 -4) = $result2 and Result3 (7 + 8)* 9/3 -4 = $result3 "
?>
[#7] janturon at email dot cz [2007-10-08 06:42:35]
This is very common problem: set one variable to another, if it is not empty. If it is, set it to something else.
For example: set $bar to $foo, if $foo is empty, set $bar to "undefined";
if(!empty($foo)) $bar= $foo; else $bar= "undefined";
OR operator can shorten it:
$bar= @$foo or $bar= "undefined";
[#8] me at robrosenbaum dot com [2007-07-12 12:16:03]
The scope resolution operator ::, which is missing from the list above, has higher precedence than [], and lower precedence than 'new'. This means that self::$array[$var] works as expected.
[#9] madcoder at gmail dot com [2007-06-09 15:17:29]
In response to mathiasrav at gmail dot com:
The reason for that behavior is the parentheses. From the description:
"Parentheses may be used to force precedence, if necessary. For instance: (1 + 5) * 3 evaluates to 18."
So the order of operations says that even though the equality operator has higher precedence, the parentheses in your statement force the assignment operator to a higher precedence than the equality operator.
That said, it still doesn't work the way you expect it to. Neither way works, for these reasons:
<?php
if ( $a != ($a = $b) )
?>
Order of operations says to do the parentheses first. So you end up with:
<?php
$a = $b;
if ( $a != $a )
?>
Which is obviously going to be false. Without the parentheses:
<?php
if ( $a != $a = $b )
?>
Order of operations says to do the inequality first, then the assignment, so you have:
<?php
if ( $a != $a );
$a = $b;
?>
Which again is not what you expected, and again will always be false. But because you are only working with values of 0 and 1, you can make use of the XOR operator:
<?php
if ( $a ^= $b )
?>
This will only be true if 1) $a is 0 and $b is 1, or 2) $a is 1 and $b is 0. That is precisely what you wanted, and it even does the assignment the way you expected it to.
<?php
foreach ($ourstring as $c) {
if ($bold ^= $c['bold']) $resstring .= bold;
if ($underline ^= $c['underline']) $resstring .= underline;
$resstring .= $c[0];
}
?>
That code now works and produces the output you expected.
[#10] golotyuk at gmail dot com [2006-07-09 09:51:38]
Simple POST and PRE incremnt sample:
<?php
$b = 5;
$a = ( ( ++$b ) > 5 ); // Pre-increment test
echo (int)$a;
$b = 5;
$a = ( ( $b++ ) > 5 ); // Post-increment test
echo (int)$a;
?>
This will output 10, because of the difference in post- and pre-increment operations
[#11] rick at nomorespam dot fourfront dot ltd dot uk [2005-09-02 03:51:21]
A quick note to any C developers out there, assignment expressions are not interpreted as you may expect - take the following code ;-
<?php
$a=array(1,2,3);
$b=array(4,5,6);
$c=1;
$a[$c++]=$b[$c++];
print_r( $a ) ;
?>
This will output;-
Array ( [0] => 1 [1] => 6 [2] => 3 )
as if the code said;-
$a[1]=$b[2];
Under a C compiler the result is;-
Array ( [0] => 1 [1] => 5 [2] => 3 )
as if the code said;-
$a[1]=$b[1];
It would appear that in php the increment in the left side of the assignment is processed prior to processing the right side of the assignment, whereas in C, neither increment occurs until after the assignment.
[#12] [2004-06-09 17:58:07]
of course this should be clear, but i think it has to be mentioned espacially:
AND is not the same like &&
for example:
<?php $a && $b || $c; ?#=#>
is not the same like
<?php $a AND $b || $c; ?#=#>
the first thing is
(a and b) or c
the second
a and (b or c)
'cause || has got a higher priority than and, but less than &&
of course, using always [ && and || ] or [ AND and OR ] would be okay, but than you should at least respect the following:
<?php $a = $b && $c; ?>
<?php $a = $b AND $c; ?>
the first code will set $a to the result of the comparison $b with $c, both have to be true, while the second code line will set $a like $b and THAN - after that - compare the success of this with the value of $c
maybe usefull for some tricky coding and helpfull to prevent bugs :D
greetz, Warhog
[#13] yasuo_ohgaki at hotmail dot com [2001-03-25 23:53:28]
Other Language books' operator precedence section usually include "(" and ")" - with exception of a Perl book that I have. (In PHP "{" and "}" should also be considered also). However, PHP Manual is not listed "(" and ")" in precedence list. It looks like "(" and ")" has higher precedence as it should be.
Note: If you write following code, you would need "()" to get expected value.
<?php
$bar = true;
$str = "TEST". ($bar ? 'true' : 'false') ."TEST";
?>
Without "(" and ")" you will get only "true" in $str.
(PHP4.0.4pl1/Apache DSO/Linux, PHP4.0.5RC1/Apache DSO/W2K Server)
It's due to precedence, probably.