PHPStan :如何返回子类型，而不是父类型

Question

我花了很多时间阅读PHPStan文档和尝试，但是我不明白如何让PHPStan在这段代码中检测到子类，甚至可能.

<?php declare(strict_types = 1);

abstract class Bar {}
    
class Foo extends Bar 
{
}
    
class Foofoo extends Bar 
{
}
    
class Baz
{
    private Bar $_bar;
        
    /**
     * @param Bar $bar
     *
     * @return void
     */
    public function setBar(Bar $bar)
    {
        $this->_bar=$bar;
    }
    
    /**
     *
     * @return Bar
     */
    public function bar(): Bar
    {
        return $this->_bar;
    }
}
   
$baz=new Baz();
$baz->setBar(new Foo());
\PHPStan\dumpType($baz->bar());     // return type is Bar, I would like Foo
$baz->setBar(new Foofoo());
\PHPStan\dumpType($baz->bar());     // return type is Bar, I would like Foofoo

到操场的链接：https://phpstan.org/r/53aebcc3-717b-445c-8715-0f277151270b

谢谢你的帮助！

Answer 1

根据PHPStan的创始人(参见Github讨论)：

实现这一目标的唯一方法是使用泛型：https://phpstan.org/r/5ed2bf49-e9fe-43b6-b279-182458ff358a

<?php declare(strict_types = 1);

abstract class Bar {
}

class Foo extends Bar 
{
}

class Foofoo extends Bar 
{
}

/** @template T of Bar */
class Baz {
    private Bar $_bar;

    /**
     * @param T $bar
     *
     * @return void
     */
    public function setBar(Bar $bar)
    {
        $this->_bar=$bar;
    }
    
    /**
    *
    * @return T
    */
    public function bar(): Bar
    {
        return $this->_bar;
    }
}

/** @var Baz<Foo> */
$baz=new Baz();
$baz->setBar(new Foo());
\PHPStan\dumpType($baz->bar());     // return type is Bar, I would like Foo
$baz->setBar(new Foofoo());
\PHPStan\dumpType($baz->bar());     // return type is Bar, I would like Foofoo