如何在PHP中强制实现类型提示接口的契约

<?php 
namespace AppSample;

interface A
{
    public function doSomething();
}

和B类实现接口A.

<?php
namespace AppSample;

class B implements A
{
    public function doSomething()
    {
        //do something
    }

    public function doBOnlyThing()
    {
        //do thing that specific to B
    }  
}

C类将取决于接口A.

<?php
namespace AppSample;

class C
{
    private $a;

    public function __construct(A $a)
    {
        $this->a = $a;
    }

    public function doManyThing()
    {
        //this call is OK
        $this->a->doSomething();

        //if $this->a is instance of B,//PHP does allow following call
        //how to prevent this?
        $this->a->doBOnlyThing();            
    }  
}

...
(new C(new B()))->doManyThing();

如果实例类B被传递给C,那么PHP确实允许调用B的任何公共方法,即使我们只输入构造函数来接受A接口.

如何在PHP的帮助下防止这种情况,而不是依赖任何团队成员来遵守接口规范？

更新：让我们假设我不能将doBOnlyThing()方法设为私有,因为它在其他地方是必需的,或者它是我无法更改的第三方库的一部分.