ArrayObject::__construct()

(PHP 5, PHP 7)

Construct a new array object

说明

This constructs a new array object.

参数

The$inputparameter accepts an array or an Object.

Flags to control the behaviour of the ArrayObject object. See ArrayObject::setFlags().

Specify the class that will be used for iteration of the ArrayObject object.

返回值

Returns an ArrayObject object on success.

错误／异常

Throws InvalidArgumentException when:

• $inputis not an array or object
• $flagsis not an integer
• $iterator_classis not an object that implements Iterator

范例

ArrayObject::__construct() example

以上例程会输出：

object(ArrayObject)#1 (3) {
  [1]=>
  string(3) "one"
  [2]=>
  string(3) "two"
  [3]=>
  string(5) "three"
}

参见

• ArrayObject::setflags() Sets the behavior flags

BTW, if you need to change array later, use exchangeArray() method. Good to know when you are writing a class that extends ArrayObject()
AFAIK, exchangeArray() doesn't return anything. 

The great confusion with this class is in its naming. ArrayObject infers it will behave as an Array and as an Object. It won't. It behaves as an array. It would better be called ArrayType. You can, with some work, get it to work both as an object and as an array, but that is up to you.

As Marcus explained, the flag ArrayObject::SPL_ARRAY_AS_PROPS means the array element may be used as a property if there is no conflict with visible properties.
If there are visible properties in the class, the array element will not overwrite it's value.

Result: 1

Result: 2

Note that the first argument to ArrayObject::__construct, the initial array, is passed by reference. Nevertheless, modification of the array doesn't modify the object, so it may cause unexpected behaviour.

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