數組：如何返回參考/指針

Question

大家好！

如何返回函數中的數組引用/指針？

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); 

謝謝！

Answer 1

也許是這樣的：

$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 2

試試這個

<?php 
$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); 
?> 

Answer 3

當你通過引用傳遞使用相同的變量函數內部，你做的外面，你不回一個值。 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); 

如果這是你所有的功能是做那麼我會建議只設置這些嵌套陣列正常。

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

Answer 4

function &getRef(&$ref){ 

return $ref['test']; 

通過使用在函數開始的符號，你返回varible而不是價值的REFFERENCE。