PHP如何通过引用传递数组

Question

1. What is a correct way to use array_splice in PHP? The function header clearly says:

   array_splice ( array &$input , int $offset... so it should accept reference as a first argument.

   However, a line

   array_push(&$this->contextsIds, $contextId);

   Triggers an error Deprecated: Call-time pass-by-reference has been deprecated in ... line 132

2. How do I return a reference to an array? I have:

   public function &getContextsIds() {
       return is_array($this->contextsIds) ? $this->contextsIds : array();    
   }
   

   but it says Notice: Only variable references should be returned by reference

- tillda

Answer 1

1. The function is already declared to take a reference (array &$input); you don't need the & again when calling the function. Just call it like this:

   array_push($this->contextsIds, $contextId);
   

2. As the message says, you should only return actual variables by reference, not mere values. In your example, there are two such instances: the ? : operator evaluates to a value, and also array() by itself is just a value not bound to any variable. You should probably just return your class member regardless of whether it is empty or not:

   return $this->contextIds;
   

- casablanca

---

谢谢！你能帮我处理我添加的第二个问题吗？ - tillda

无论它是否为空，他正在测试 is_array()。 - BoltClock

@BoltClock：我注意到了，但我猜他可能是在测试它是否为空。我会等待原帖作者的回复。 - casablanca

Answer 2

为什么你会返回一个数组的引用，尤其是在你提供的代码中？

public function &getContextsIds() {
    return is_array($this->contextsIds) ? $this->contextsIds : array();    
}

当该函数工作时，它可以返回一个空数组的引用，我可以随意更改它，但它不会产生任何影响，因为它只是一个没有任何进一步引用的空数组。

- GolezTrol