错误，而试图在PHP

Question

养活一类参数i声明的类是这样的：在主代码

class Foo{ public function __construct(){ echo 'Foo was created!';} } 

class Foo2 extends Foo{ public function __construct(){ parent::__construct(); echo 'Foo2 was created!';} } 

class Bar{ 
    public function __construct(Foo $foo){ echo 'Bar was created!';} 
} 

：

$foo2 = new Foo2(); 
$bar = new Bar($foo2); 

这是什么错误主代码的原因：

Fatal error: Default value for parameters with a class type hint can only be NULL 

php版本：PHP 5.3.2

--------------------------------------更新！--------- -------------------------------

文件：system.data.php

namespace system\data{ 

    include ('system.php'); 

    use system;  
class DBConnection implements system\IDisposable { 
    protected $serverName; 
    protected $userId; 
    protected $password; 
    protected $handler; 
    protected $isOpened; 

    /* 
    * create a new instance of DBConnection. 
    */ 
    public function __construct($server, $uid, $password) { 
     $this->isOpened = false; 
     $this->serverName = $server; 
     $this->userId = $uid; 
     $this->password = $password; 
    } 
class DBCommand implements \system\IDisposable { 

    public function __construct(DBConnection $connection, int $type) { 
     $this->connection = $connection; 
     $this->queryType= $type; 
} 

}

文件：system.data.mysql.php

namespace system\data\mysql{ 
    class MySqlCommand extends DBCommand { 

     public function __construct(data\DBConnection $connection, int $type = 0) { 
      parent::__construct($connection, $type); 
     } 
    } 

class MySqlConnection extends DBConnection { 

    public function __construct($server, $uid, $password) { 
     parent::__construct($server, $uid, $password); 
     } 

    } 
} 

错误：

Fatal error: Default value for parameters with a class type hint can only be NULL in C:\Program Files\Apache Software Foundation\...\system.data.mysql.php on line 35(constructor declartion) 

Answer 1

谢谢你的完整的代码，这里是你的错误：

public function __construct(data\DBConnection $connection, int $type = 0) 

PHP 5.3只支持类型提示的阵列和类。

你的int $type = 0声明寻找类命名为“INT”，并分析它是这样，即使该类不存在。

您需要删除虚假类型提示。 PHP开发人员是considering adding other type hints in the future。有currently code in PHP's trunk为标量值执行类型提示，但自从发布该博客文章以来，发生了很多变化。我会看看我是否可以追踪目前的状况，但是搜索PHP内部邮件列表是一件非常可怕的任务。

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编辑：通过PHP内部挖掘后，我发现the latest Type Hinting线程。它看起来像this the current Type Hinting RFC，但在方向上似乎没有达成共识。 Zeev想要回滚提交，但尚不清楚这是否完成。