根据PHP文档，它是按值计算的。

默认情况下，函数实参是按值传递的(这样，如果函数内实参的值改变了，它不会在函数外被改变)。为了允许函数修改其参数，它们必须通过引用传递。 要让函数的实参始终通过引用传递，请在函数定义中的实参名称前加上&。

<?php
function add_some_extra(&$string)
{
    $string .= 'and something extra.';
}

$str = 'This is a string, ';
add_some_extra($str);
echo $str;    // outputs 'This is a string, and something extra.'
?>

两种方法都可以。

在前面放一个“&”符号，你传递的变量就变成了它的原点，也就是说，你可以通过引用传递，而不是复制它。

so

    $fred = 5;
    $larry = & $fred;
    $larry = 8;
    echo $fred;//this will output 8, as larry and fred are now the same reference.

PHP引用是一个别名，允许两个不同的变量写入相同的值。

在PHP中，如果你有一个包含对象的变量，这个变量不包含对象本身。相反，它包含该对象的标识符。对象访问器将使用标识符查找实际对象。因此，当我们在函数中使用对象作为参数或将其赋值给另一个变量时，我们将复制指向对象本身的标识符。

https://hsalem.com/posts/you-think-you-know-php.html

class Type {}

$x = new Type();
$y = $x;
$y = "New value";

var_dump($x); // Will print the object.
var_dump($y); // Will print the "New value"

$z = &$x; // $z is a reference of $x

$z = "New value";
var_dump($x); // Will print "New value"
var_dump($z); // Will print "New value"

可以通过引用将变量传递给函数。这个函数将能够修改原始变量。

你可以在函数定义中通过引用来定义段落:

<?php
function changeValue(&$var)
{
    $var++;
}

$result=5;
changeValue($result);

echo $result; // $result is 6 here
?>

在PHP 5中通过引用传递对象，在PHP 4中通过值传递对象。 默认情况下，变量是按值传递的!

阅读此处:http://www.webeks.net/programming/php/ampersand-operator-used-for-assigning-reference.html

TL;DR: PHP supports both pass by value and pass by reference. References are declared using an ampersand (&); this is very similar to how C++ does it. When the formal parameter of a function is not declared with an ampersand (i.e., it's not a reference), everything is passed by value, including objects. There is no distinction between how objects and primitives are passed around. The key is to understand what gets passed along when you pass in objects to a function. This is where understanding pointers is invaluable.

对于将来遇到这种情况的人，我想分享这个来自PHP文档的精华，由一位匿名用户发布:

这里似乎有些混乱。指针和引用之间的区别并不是特别有用。 已经发布的一些“综合”示例中的行为可以用更简单的统一术语来解释。例如，Hayley的代码正在做您应该期望它应该做的事情。(使用>= 5.3)

第一个原则: 指针存储访问对象的内存地址。任何时候赋值一个对象，都会生成一个指针。(我还没有深入研究Zend引擎，但据我所知，这是适用的)

2nd principle, and source of the most confusion: Passing a variable to a function is done by default as a value pass, ie, you are working with a copy. "But objects are passed by reference!" A common misconception both here and in the Java world. I never said a copy OF WHAT. The default passing is done by value. Always. WHAT is being copied and passed, however, is the pointer. When using the "->", you will of course be accessing the same internals as the original variable in the caller function. Just using "=" will only play with copies.

3rd principle: "&" automatically and permanently sets another variable name/pointer to the same memory address as something else until you decouple them. It is correct to use the term "alias" here. Think of it as joining two pointers at the hip until forcibly separated with "unset()". This functionality exists both in the same scope and when an argument is passed to a function. Often the passed argument is called a "reference," due to certain distinctions between "passing by value" and "passing by reference" that were clearer in C and C++.

只要记住:传递给函数的是指向对象的指针，而不是对象本身。这些指针是原始指针的副本，除非在参数列表中使用“&”来实际传递原始指针。只有当你深入到一个物体的内部时，原始的东西才会改变。

下面是他们提供的例子:

<?php

//The two are meant to be the same
$a = "Clark Kent"; //a==Clark Kent
$b = &$a; //The two will now share the same fate.

$b="Superman"; // $a=="Superman" too.
echo $a;
echo $a="Clark Kent"; // $b=="Clark Kent" too.
unset($b); // $b divorced from $a
$b="Bizarro";
echo $a; // $a=="Clark Kent" still, since $b is a free agent pointer now.

//The two are NOT meant to be the same.
$c="King";
$d="Pretender to the Throne";
echo $c."\n"; // $c=="King"
echo $d."\n"; // $d=="Pretender to the Throne"
swapByValue($c, $d);
echo $c."\n"; // $c=="King"
echo $d."\n"; // $d=="Pretender to the Throne"
swapByRef($c, $d);
echo $c."\n"; // $c=="Pretender to the Throne"
echo $d."\n"; // $d=="King"

function swapByValue($x, $y){
$temp=$x;
$x=$y;
$y=$temp;
//All this beautiful work will disappear
//because it was done on COPIES of pointers.
//The originals pointers still point as they did.
}

function swapByRef(&$x, &$y){
$temp=$x;
$x=$y;
$y=$temp;
//Note the parameter list: now we switched 'em REAL good.
}

?>

关于JavaScript的这个主题，我写了一篇广泛而详细的博客文章，但我相信它同样适用于PHP、c++和任何其他人们似乎对按值传递和按引用传递感到困惑的语言。

Clearly, PHP, like C++, is a language that does support pass by reference. By default, objects are passed by value. When working with variables that store objects, it helps to see those variables as pointers (because that is fundamentally what they are, at the assembly level). If you pass a pointer by value, you can still "trace" the pointer and modify the properties of the object being pointed to. What you cannot do is have it point to a different object. Only if you explicitly declare a parameter as being passed by reference will you be able to do that.