How do I add an array as an Object Property to a class declared within a PHP extension?

Question

I want my PHP extension to declare a class equivalent to the following PHP:

class MyClass
{
    public $MyMemberArray;

    __construct()
    {
        $this->MyMemberArray = array();
    }
}

I'm following the examples in "Advanced PHP Programming" and "Extending and Embedding PHP" and I'm able to declare a class that has integer properties in PHP_MINIT_FUNCTION.

However, when I use the same approach to declare an array property in PHP_MINIT_FUNCTION, I get the following error message at runtime:

PHP Fatal error:  Internal zval's can't be arrays, objects or resources in Unknown on line 0

There's an example on page 557 of Advanced PHP Programming of how to declare a constructor which creates an array property, but the example code doesn't compile (the second "object" seems to be redundant).

I fixed the bug and adapted it to my code:

PHP_METHOD(MyClass, __construct)
{
    zval *myarray;
    zval *pThis;

    pThis = getThis();

    MAKE_STD_ZVAL(myarray);

    array_init(myarray);
    zend_declare_property(Z_OBJCE_P(pThis), "MyMemberArray", sizeof("MyMemberArray"), myarray, ZEND_ACC_PUBLIC TSRMLS_DC);
}

And this compiles, but it gives the same runtime error on construction.

John Carter · Accepted Answer · 2009-07-09 19:14:15Z

6

The answer is to use add_property_zval_ex() in the constructor, rather than zend_declare_property().

The following works as intended:

PHP_METHOD(MyClass, __construct)
{
    zval *myarray;
    zval *pThis;

    pThis = getThis();

    MAKE_STD_ZVAL(myarray);

    array_init(myarray);
    add_property_zval_ex(pThis, "MyMemberArray", sizeof("MyMemberArray"), myarray);
}

edited Jul 9, 2009 at 19:14

answered Jul 9, 2009 at 18:04

John Carter

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2 Comments

Christopher Thomas Over a year ago

but do you know WHY this is the answer? what is the reasoning?

markzz Over a year ago

The reasoning has to do with how a zval is created. It is a long winded explaination, trust me.

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