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When Changing the Value of Object Variables, How Does PHP Handle Assignment of Reference and Value?

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2024-10-22 07:22:30439browse

When Changing the Value of Object Variables, How Does PHP Handle Assignment of Reference and Value?

Assignment in PHP: Passing by Value or Reference

Objects in PHP are handled differently than primitive data types when it comes to assignment. By default, PHP assigns variables by reference, meaning that changing the value of an object's variable will also affect the original object. This behavior can be somewhat confusing, especially when working with arrays of objects.

Code Explanation

Consider the following code snippet:

<code class="php">class Foo
{
    // ...
}

class Bar
{
    private array $foos;

    // ...

    public function getFoo(int $index): Foo
    {
        return $this->foos[$index];
    }

    public function test(): void
    {
        $testFoo = $this->getFoo(5);
        $testFoo->setValue("My value has now changed");
    }
}

$b = new Bar;
echo $b->getFoo(5)->value;
$b->test();
echo $b->getFoo(5)->value;</code>

Passing by Value and By Reference

Now, let's analyze the code and answer the question:

When the method Bar::test() is run and it changes the value of foo #5 in the array of foo objects, will the actual foo #5 in the array be affected?

When the getFoo method returns the Foo object at index 5, it returns a copy of the object by default since PHP assigns variables by value for primitive data types and copies for reference data types by default. Therefore, the changes made to the $testFoo object won't affect the original object in the array until it's assigned back to the array.

Assignment by Reference

However, the code inside test directly modifies the returned Foo object. When an assignment to an object is made, PHP assigns by reference. This means that changes made to the assigned object will be reflected in the original object. As a result, after the setValue method is called on $testFoo, the original Foo object at index 5 in the $foos array is updated.

Demonstration

Running the code will produce the following output:

Foo #5
My value has now changed

This output demonstrates that the assignment by reference behavior has affected the original object in the array.

Changing Assignment Behavior

If you want to assign an object by value (creating a copy) instead of by reference, you can use the 'clone' keyword, as shown below:

<code class="php">$testFoo = clone $this->getFoo(5);</code>

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