public方法访问的有关问题

WBOY
Release: 2016-06-13 13:12:52
Original
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public方法访问的问题。
  class A{
public function pp(){
return 1300;
  }
public function __construct(){
echo 1;
}
  }
  class B{
  public function __construct(){
//echo A::__construct();
//echo A::pp();
}
  }
  $cc=new C;
?>

问个问题:
pp是class A的公有方法,所以可以用类名+双冒号的形式访问。那问题就在于,为什么class A里头的__construct方法,我也是给了public,而在class B里头,去A::__construct()调用的时候会报以下错误:

Fatal error: Non-static method A::__construct() cannot be called statically, assuming $this from incompatible context in D:\wamp\www\1.php on line 12

还有个问题,比如:[b][/b]
class A{
  public $m=10;
}
这个$m既然是public,那就应该支持类内部访问,子类访问以及实例访问。
我在有些书上看到,它们说“在任何地方”都可以访问。我想问这个任何地方是否包括其他不相关的类(和A没继承关系的类),如果支持访问,怎么访问?代码如何写的?



------解决方案--------------------
静态方法或者实例化

你上面的代码实例化了 例如这样访问

PHP code

class A{
    public $mm=10;
}
class B{
    protected $classA;
    public function __construct($number){
        $this->classA=$number;
        echo $this->classA;
    }
}
$aa=new A;
$bb=new B($aa->mm); <div class="clear">
                 
              
              
        
            </div>
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