PHP变量是按值传递还是按引用传递?


当前回答

取决于版本,4是值,5是引用。

其他回答

关于如何将对象传递给函数,你仍然需要理解,没有“&”,你传递给函数的是一个对象句柄,对象句柄仍然是通过值传递的,它包含一个指针的值。但你不能改变这个指针,直到你通过引用传递它使用“&”

<?php
        class Example 
        {
            public $value;
         
        }
        
        function test1($x) 
        {
             //let's say $x is 0x34313131
             $x->value = 1;  //will reflect outsite of this function
                             //php use pointer 0x34313131 and search for the 
                             //address of 'value' and change it to 1

        }
        
        function test2($x) 
        {
             //$x is 0x34313131
             $x = new Example;
             //now $x is 0x88888888
             //this will NOT reflect outside of this function 
             //you need to rewrite it as "test2(&$x)"
             $x->value = 1000; //this is 1000 JUST inside this function
                 
        
        }
         
     $example = new Example;
    
     $example->value = 0;
    
     test1($example); // $example->value changed to  1
    
     test2($example); // $example did NOT changed to a new object 
                      // $example->value is still 1
     
 ?>

在PHP5中,包含基本类型的变量是按值传递的。包含对象的变量是通过引用传递的。2006年的Linux Journal上有一篇非常有趣的文章提到了4和5之间的OO差异。

http://www.linuxjournal.com/article/9170

在PHP中,默认情况下,对象作为新对象的引用传递。

请看这个例子:

class X {
  var $abc = 10; 
}

class Y {

  var $abc = 20; 
  function changeValue($obj)
  {
   $obj->abc = 30;
  }
}

$x = new X();
$y = new Y();

echo $x->abc; //outputs 10
$y->changeValue($x);
echo $x->abc; //outputs 30

现在看这个:

class X {
  var $abc = 10; 
}

class Y {

  var $abc = 20; 
  function changeValue($obj)
  {
    $obj = new Y();
  }
}

$x = new X();
$y = new Y();

echo $x->abc; //outputs 10
$y->changeValue($x);
echo $x->abc; //outputs 10 not 20 same as java does.

现在看这个:

class X {
  var $abc = 10; 
}

class Y {

  var $abc = 20; 
  function changeValue(&$obj)
  {
    $obj = new Y();
  }
}

$x = new X();
$y = new Y();

echo $x->abc; //outputs 10
$y->changeValue($x);
echo $x->abc; //outputs 20 not possible in java.

我希望你能理解。

http://www.php.net/manual/en/migration5.oop.php

In PHP 5 there is a new Object Model. PHP's handling of objects has been completely rewritten, allowing for better performance and more features. In previous versions of PHP, objects were handled like primitive types (for instance integers and strings). The drawback of this method was that semantically the whole object was copied when a variable was assigned, or passed as a parameter to a method. In the new approach, objects are referenced by handle, and not by value (one can think of a handle as an object's identifier).

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.'
?>