类型错误：返回值必须是实例，返回null

I am using type hinting of php7. So I have following code

class Uni
{
 /**
     * @var Student
     *
     * @ORM\ManyToOne(targetEntity="AppBundle\Entity\Student")
     * @ORM\JoinColumn(nullable=false)
     */
    private $student;

    function _construct(){
        $this->student= new Student();
    }

    /**
     * Set student
     *
     * @param \AppBundle\Entity\Student $student
     *
     * @return Uni
     */
    public function setStudent (Student $student): Uni
    {
        $this->student= $student;

        return $this;
    }

    /**
     * Get student
     *
     * @return \AppBundle\Entity\Student
     */
    public function getStudent(): Student 
    {
        return $this->student;
    }

}

now when I try to load new form for Uni, I get this error

Type error: Return value of AppBundle\Entity\Uni::getStudent() must be an instance of AppBundle\Entity\Student, null returned 

How can I get rid of this error? I found a solution with nullable and it requires php 7.1. But for now I have to stick with php 7.0. So how can I solve this?

You have a typo in your constructor, two underscores must come before construct.

function __construct() {
    $this->student = new Student();
}

Because of this, $student was null when your object was created.