如何将参数传递给尚未声明的函数？

Question

这是我的代码：

class myclass{

    public function index(){
        $words  = ['word1', 'word2', 'word3'];
        $result = sizeof($words);
        $condition = in_array($word, $words) || strlen($word) <= 3;
        $res = $this->investigate_words( $words, $result, $condition )
    }

    public function investigate_words($words, $result, $condition)
    {
        foreach($words as $word){
            if($condition){
                $result--;
            }
        }
        return $result;
    }
}

请关注这一行：

$condition = in_array($word, $words) || strlen($word) <= 3;

在这一行中，$word 尚未声明。它应该是存在于investigate_words() 函数中的循环中的值。无论如何，有什么解决办法吗？

Answer 1

在这种情况下，我们可以使用Closure。条件参数可以是可调用的。

class myclass{

    public function index(){
        $words  = ['word1', 'word2', 'word3'];
        $result = sizeof($words);
        $res = $this->investigate_words( $words, $result, function($word){
            return strlen($word) <= 3;// || in_array($word, $words); You don't need this commented condition word's coming from array
        });
    }

    public function investigate_words($words, $result,callable $condition)
    {
        foreach($words as $word){
            if($condition($word)){
                $result--;
            }
        }
        return $result;
    }
}