如何将数组作为Object属性添加到PHP扩展中声明的类？

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

我希望我的PHP扩展声明一个等效于以下PHP的类:

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.

我正在按照“高级PHP编程”和“扩展和嵌入PHP”中的示例进行操作,并且我能够在PHP_MINIT_FUNCTION中声明一个具有整数属性的类。

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

但是,当我使用相同的方法在PHP_MINIT_FUNCTION中声明一个数组属性时,我在运行时收到以下错误消息:

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).

高级PHP编程的第557页有一个例子,说明如何声明一个创建数组属性的构造函数,但是示例代码没有编译(第二个“对象”似乎是多余的)。

I fixed the bug and adapted it to my code:

我修复了错误并将其修改为我的代码:

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.

这会编译,但它在构造时会产生相同的运行时错误。

1 个解决方案

#1

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

答案是在构造函数中使用add_property_zval_ex(),而不是zend_declare_property()。

The following works as intended:

以下按预期工作:

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);
}

#1