php function 按引用返回一个值

    按引用返回一个值,而不是按值返回,这样就无需为变量建立一个重复的副本

 1 function &array_find_value($needle, &$haystack)
 2 {
 3     foreach ($haystack as $key=>$value){
 4         if ($needle == $value){
 5             //必须返回一个变量的引用,而不能是包含这个变量的字符串
 6             return $haystack[$key];
 7         }
 8     }
 9 }
10 $artists = ['table','The Doors'];
11 $band =& array_find_value('The Doors',$artists);
12 var_dump($band); // string 'The Doors'
13 //改变数组
14 $band = 'thhe doors name is panpan';
15 var_dump($artists);
16 /*
17   array (size=2)
18     0 => string 'table' (length=5)
19     1 => &string 'thhe doors name is panpan' (length=25)
20  */
原文地址:https://www.cnblogs.com/loveyouyou616/p/5541521.html