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如何复制一个没有引用的对象?

pass-by-reference object php

acm · 技术社区 · 14 年前

有很好的文件证明PHP5 OOP objects are passed by reference 默认情况下。如果这是默认的,在我看来有一个没有默认的方式复制没有参考,如何??

function refObj($object){
    foreach($object as &$o){
        $o = 'this will change to ' . $o;
    }

    return $object;
}

$obj = new StdClass;
$obj->x = 'x';
$obj->y = 'y';

$x = $obj;

print_r($x)
// object(stdClass)#1 (3) {
//   ["x"]=> string(1) "x"
//   ["y"]=> string(1) "y"
// }

// $obj = refObj($obj); // no need to do this because
refObj($obj); // $obj is passed by reference

print_r($x)
// object(stdClass)#1 (3) {
//   ["x"]=> string(1) "this will change to x"
//   ["y"]=> string(1) "this will change to y"
// }

$x 原汁原味 $obj like this

2 回复 | 直到 7 年前

Treffynnon 14 年前

<?php
$x = clone($obj);

所以应该是这样的:

<?php
function refObj($object){
    foreach($object as &$o){
        $o = 'this will change to ' . $o;
    }

    return $object;
}

$obj = new StdClass;
$obj->x = 'x';
$obj->y = 'y';

$x = clone($obj);

print_r($x)

refObj($obj); // $obj is passed by reference

print_r($x)