Php, typed property and lazy loading? How can I check if it's initialized?

Question

you know Php 7.2 has typed properties:

class Test { }

class Test2
{
    private Test $test;

    public function __construct()
    {
        $this->test = new Test();
    }
}

so far so good, but what if I want to have a lazy created object?

public function createIfNotExists()
{
    if ($this->test === null) // *ERROR
    {
    }
}

this fails:

Typed property must not be accessed before initialization

but I want to check either it's been created, not using it. How to?

isset() or !empty() ? & Use the null operator on your property: private ?Test $test = null; — Jaquarh
– Jaquarh, Commented May 4, 2021 at 21:18
If you always initialize it in the constructor, the test will never fail. — Barmar
– Barmar, Commented May 4, 2021 at 21:19
I think if (property_exists($this, "test")) is more appropriate, but Barmar is correct that you should just set it in the constructor. — miken32
– miken32, Commented May 4, 2021 at 21:30

Alberto · Accepted Answer · 2021-05-04 21:19:20Z

3

you can probably do

isset($this->test)

answered May 4, 2021 at 21:19

Alberto

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2 Comments

great, this works. I also tried to use $this->test === null, also failed.

yes because to use === you have to access the property @JohnSmith

Cid · Accepted Answer · 2021-05-04 21:30:01Z

1

As of PHP 7.4, you can use the Null Coalescing Assignment Operator ??= as well in your method

public function createIfNotExists()
{
    $this->test ??= new Test();
}

It's equivalent to

$this->test = $this->test ?? new Test();

Which is equivalent to

$this->test = isset($this->test) ? $this->test : new Test();

answered May 4, 2021 at 21:30

Cid

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