Question

Hy大家！

如何在函数中返回数组引用/指针？

例如：

$a=array('given'=>array());

function getRef(&$ref){

//adds a child element to the given reference/pointer
$ref['test']=array();

//doesn't return the current reference/pointer
return $ref['test'];
}

//out: Array ( [given] => Array ( [test] => Array ( ) ) )
$p=getRef($a['given']);
print_r($a);

//out: same as above
//expected: ( [given] => Array ( [test] => Array ([test2] => Array ( ) ) ) )
$p['test2']=array();
print_r($a);

谢谢！

Answer 1

function &getRef(&$ref){

return $ref['test'];

通过在函数开头使用＆符号，可以返回变量的refference，而不是值。

Answer 2

也许是这样的：

$a=array('given'=>array());

function getRef(&$ref){

//adds a child element to the given reference/pointer
$ref['test']=array();

//doesn't return the current reference/pointer
return $ref['test'];
}

//out: Array ( [given] => Array ( [test] => Array ( ) ) )
$p=getRef($a['given']);
print_r($a);

//out: same as above
//expected: ( [given] => Array ( [test] => Array ([test2] => Array ( ) ) ) )
$p['given']['test']['test2']=array();
print_r($p);

Answer 3

当您通过引用传递时，您在函数内部使用与外部相同的变量，并且您不会返回值。 PHP Manual - Passing by Reference

$a = array('given' => array());

function getRef(&$a)
{
  $a['test']=array();   
}

// this now has the nested array test.
getRef($a['given']);
print_r($a);

如果这是您的所有功能，那么我建议您正常设置这些嵌套数组。

$b = ['given' => ['test' => ['test2' => []]]];
print_r($b);

array：如何返回引用/指针

3 个答案: