php - 如何复制 ArrayIterator 以保留其当前迭代位置？

Question

因为这似乎是我必须做的才能获得这种效果:

$arr = ['a'=>'first', 'b'=>'second', ...];
$iter = new ArrayIterator( $arr );

// Do a bunch of iterations...
$iter->next();
// ...

$new_iter = new ArrayIterator( $arr );
while( $new_iter->key() != $iter->key() ) {
    $new_iter->next();
}

编辑:此外，为了清楚起见，我不应该使用 unset() 修改基本数组吗？我认为数组迭代器存储它自己的基本数组副本，因此使用 offsetUnset() 似乎不正确。

Answer 1

ArrayIterator没有实现 tell() 函数，但您可以模拟它，然后使用 seek()去你想要的位置。这是一个扩展类，它就是这样做的:

<?php
    class ArrayIteratorTellable extends ArrayIterator {
        private $position = 0;

        public function next() {
            $this->position++;
            parent::next();
        }

        public function rewind() {
            $this->position = 0;
            parent::rewind();
        }

        public function seek($position) {
            $this->position = $position;
            parent::seek($position);
        }

        public function tell() {
            return $this->position;
        }

        public function copy() {
            $clone = clone $this;
            $clone->seek($this->tell());
            return $clone;
        }
    }
?>

使用:

<?php
    $arr = array('a' => 'first', 'b' => 'second', 'c' => 'third', 'd' => 'fourth');
    $iter = new ArrayIteratorTellable( $arr );

    $iter->next();

    $new_iter = new ArrayIteratorTellable( $arr );

    var_dump($iter->current()); //string(6) "second"
    var_dump($new_iter->current()); //string(6) "first"

    $new_iter->seek($iter->tell()); //Set the pointer to the same as $iter

    var_dump($new_iter->current()); //string(6) "second"
?>

DEMO

---

或者，您可以使用自定义 copy() 函数克隆对象:

<?php
    $arr = array('a' => 'first', 'b' => 'second', 'c' => 'third', 'd' => 'fourth');
    $iter = new ArrayIteratorTellable( $arr );

    $iter->next();

    $new_iter = $iter->copy();

    var_dump($iter->current()); //string(6) "second"
    var_dump($new_iter->current()); //string(6) "second"
?>

DEMO