PHP中的数组在传递给函数时会复制为值还是对新变量的引用?

问题描述：

1)当数组作为参数传递给方法或函数时，它是通过引用还是通过值传递?

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

2)将数组分配给变量时，新变量是对原始数组的引用，还是新副本?
怎么办呢?

2) When assigning an array to a variable, is the new variable a reference to the original array, or is it new copy?
What about doing this:

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

$b是对$a的引用吗?

答

对于问题的第二部分，请参见

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

数组分配总是涉及值复制.使用引用运算符通过引用复制数组.

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

给出的示例:

<?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
?>

对于第一部分，最好的确定方法是尝试;-)

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

考虑以下示例代码:

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

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

它将给出以下输出:

array
  0 => int 10
  1 => int 20

这表明该函数尚未修改作为参数传递的外部"数组:它作为副本而不是引用传递.

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.

如果要通过引用传递它，则必须以这种方式修改函数:

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

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

输出将变为:

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

因为，这一次，该数组已通过引用"传递.

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

不要犹豫，阅读手册的参考资料部分:它应该回答您的一些问题;-)

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

PHP中的数组在传递给函数时会复制为值还是对新变量的引用?

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