PHP扩展值不正确打印原因 - php extension value not correct printed why?

php extension value not correct printed why?

本文关键字：打印不正确扩展 PHP | 更新日期: 2023-09-27

我想通过引用发送一个对象。对象表示一个类

[test.php]

$car= new Car();
$car->l=1000;
$car2 = new Car2();
$car2->method($car);

[php_cod.cc]

PHP_METHOD(Car2, method)
{
    Car2 *car;
    Car obj11;
    zend_class_entry ce2;
     if (zend_parse_parameters(ZEND_NUM_ARGS() TSRMLS_CC, "o", &obj11) == FAILURE) {
        RETURN_NULL();
    }

    car2_object *obj = (car2_object *)zend_object_store_get_object(
        getThis() TSRMLS_CC);
    car2 = obj->car2;
    if (car2 != NULL) {
        //cout<<"in car 2 ref"<<endl;
        //car2->reference(s);
        (car2->method(obj11));
    }
}

[test.cc]

void Car2::method(Car &carr)
{
    cout<<"IN THE REFERENCE CLASS"<<carr.l<<endl; 
}

我错在哪里?谢谢!欣赏

当我运行php test.php时，值将返回。L等于155160836. 为什么?我哪里错了

在php代码部分，obj11是一个未初始化的指针，它被传递给Car2::方法((car2->method(*obj11));)。所以- - - - - -

 cout<<"IN THE REFERENCE CLASS"<<carr.l<<endl; 
 // carr is a garbage reference and is causing the segmentation fault