php - 如何为对象创建默认值？ PHP

Question

假设我有一个 Human 类，它的变量 $gender 没有分配任何值。 Human 有一个带有年龄、性别、高度和体重参数的构造函数。

我有另一个名为 Female 的类，它继承自 Human，但现在 Female 类正在用字符串 Female 覆盖 $gender 变量。

当我创建对象时，假设 $f = new Female(12, 'female', 123, 40);如何在创建对象时跳过输入女性？

我认为我们需要在 Female 类中创建另一个新的构造函数，我在 Female 类构造函数的参数中有 age, gender = 'female', height and weight 但这不是似乎有效。

我尝试在创建对象时将性别部分留空或尝试输入空字符串，例如 ""。

有人可以帮我一下吗？非常感谢。

人类类代码

class Human {
    protected $age = 0;   
    protected $gender;
    protected $height_in_cm;
    protected $weight_in_kg;

    function __construct($age, $gender, $heightCM, $weightKG)
    {
        $this->age = $age;
        $this->gender = $gender;
        $this->height_in_cm = $heightCM;
        $this->weight_in_kg = $weightKG;
    }

    /**
     * @return int
     */
    public function getAge()
    {
        return $this->age;
    }

    /**
     * @return string
     */
    public function getGender()
    {
        return $this->gender;
    }
}

女性类代码

require_once('Human.php');
class Female extends Human{
    protected $gender = 'female';

    function __construct($age, $gender = 'female', $heightCM, $weightKG)
    {
        $this->age = $age;
        $this->gender = $gender;
        $this->height_in_cm = $heightCM;
        $this->weight_in_kg = $weightKG;
    }
}

$f = new Female(12,'female',123,40);
echo "Your gender is ". $f->getGender()."<br>";

Answer 1

简单地覆盖构造函数:

class Human {
     public function __construct($age, $gender, $height, $weight) {
     }
}

class Female extends Human {
    public function __construct($age, $height, $weight) {
         parent::__construct($age, 'Female', $height, $weight);
    }
}