在PHP数组按​​值或按引用传递？ [英] are arrays in php passed by value or by reference?

问题描述

当一个数组作为参数传递给方法或函数传递是按引用传递？

When an array is passed as an argument to a method or function is it passed by reference?

关于这样内容：

What about doing this:

    $a = array(1,2,3);
    $b = $a

时的$ B $到一个参考？

Is $b a reference to $a?

推荐答案

有关你问题的第二部分，看到的手动阵列页，其中规定的（报价）的：

For the second part of your question, see the array page of the manual, which states (quoting) :

数组赋值总是涉及价值
  复制。使用参考运算符
  通过参考复制的阵列

Array assignment always involves value copying. Use the reference operator to copy an array by reference.

和给定的例子：

<?php
$arr1 = array(2, 3);
$arr2 = $arr1;
$arr2[] = 4; // $arr2 is changed,
             // $arr1 is still array(2, 3)

$arr3 = &$arr1;
$arr3[] = 4; // now $arr1 and $arr3 are the same
?>

结果
在第一部分，以确保最好的办法是尝试; - ）

For the first part, the best way to be sure is to try ;-)

考虑code的这个例子：

Consider this example of code :

function my_func($a) {
    $a[] = 30;
}

$arr = array(10, 20);
my_func($arr);
var_dump($arr);

这会给这样的输出：

It'll give this output :

array
  0 => int 10
  1 => int 20

这表示该功能没有修改被作为参数传递外部阵：这是通过为副本，而不是引用

Which indicates the function has not modified the "outside" array that was passed as a parameter : it's passed as a copy, and not a reference.

如果你想要通过引用传递，你就必须修改功能，这种方式：

If you want it passed by reference, you'll have to modify the function, this way :

function my_func(& $a) {
    $a[] = 30;
}

和输出将变为：

array
  0 => int 10
  1 => int 20
  2 => int 30

由于，这个时候，阵列已通过引用传递。

As, this time, the array has been passed "by reference".

结果
不要犹豫，阅读href=\"http://php.net/manual/en/language.references.php\">引用的解释手册的部分的

Don't hesitate to read the References Explained section of the manual : it should answer some of your questions ;-)

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