在php中,如何将一个类实例传递给另一个类的构造函数

我正在尝试实现桥设计模式,遵循以下步骤 Tutorials Point . 我正在将代码从java转换为php并更改一些名称。

问题是,当我试图将具体的桥实现程序类传递到具体的类实现接口时,会抛出一个错误。

在代码中更容易理解:

// LaunchApi.php
interface LaunchApi
{
    public function launch();
}

// RedEngine.php
class RedEngine implements LaunchApi
{
    public function launch()
    {
        echo "The red engine is really fast!!!";
    }
}

// Rocket.php
abstract class Rocket
{
    protected $launchApi;

    protected function __construct($launchApiImplementer)
    {
        $this->launchApi = $launchApiImplementer;
    }

    public abstract function launch();
}

// FullRocket.php
class FullRocket extends Rocket
{
    public function __construct($launchApi)
    {
        parent::__construct($launchApi);
    }

    public function launch()
    {
        $this->launchApi->launch();
    }
}

// LaunchingScript.php
$redEngine = new RedEngine();
$redEngine->launch(); // this works

$redRocket = new FullRocket($redEngine);
$redRocket.launch(); // this won't work

错误抛出为:

design-patterns\Bridge>php LaunchingBridge.php
The red engine is really fast!!!
Fatal error: Call to undefined function launch() in \design-patterns\Bridge\LaunchingBridge.php on line 24

我试图使用&通过引用传递,但它只会更改错误。