禁止扩展方法(Forbid extension methods)

 我认为在你的情况下，你需要确保函数calculateAge能够calculateAge传递对象的年龄。 例如，为了计算对象的年龄（看起来它是一个动物:)），函数calculateAge知道无论对象的类型如何，它都有一个名为getAge的方法来完成这项工作。 因此，您需要在calculateAge执行的所有操作都是调用传递对象的getAge 。 
 
 你可以这样做，例如： 
 
interface Animal
{
    public function getAge();
}

// The same thing for the class Dog
class Cat implements Animal
{
     public function getAge()
     {
           // code to calculate age
     }
     ...
}

public function calculateAge(Animal $animal)
{
    echo $animal->getAge();
}

$dog = new Dog();
calculateAge($dog);

$cat = new Cat();
calculateAge($cat);

I think in your case you need at to make sure your function calculateAge is able to calcualte the age of the passed object. For instance, to calculate the age of the object (it seems that it's an animal :) ), your function calculateAge knows that whatever the type of the object, it has a method called getAge that does the job. So all what you need to do in calculateAge is to call getAge of the passed object.
 
You can do this for instance:
 
interface Animal
{
    public function getAge();
}

// The same thing for the class Dog
class Cat implements Animal
{
     public function getAge()
     {
           // code to calculate age
     }
     ...
}

public function calculateAge(Animal $animal)
{
    echo $animal->getAge();
}

$dog = new Dog();
calculateAge($dog);

$cat = new Cat();
calculateAge($cat);