I can't express my question in words. Please look the code below, I hope you will understand my question.
I have a class and an interface as shown below.
class MyInterface
{
public:
virtual ~MyInterface(){}
virtual void print() = 0;
};
class MyClass : public MyInterface
{
public:
MyClass(){}
~MyClass(){}
void print()
{
printf("Hello World\n");
}
};
Now here's my question.
MyClass* myclass = new MyClass();
myclass->print(); //will print "Hello World"
MyInterface* pMyInterface = (MyInterface*)myclass;
pMyInterface->print();
Will the second call print Hello World as well? If yes, then why?
One note is that you do not need to explicitly cast to an accessible base class, like you do in MyInterface* pMyInterface = (MyInterface*)myclass;. It is an implicit conversion from a pointer/reference to a derived class to that of an accessible base class.
In fact, such casting may introduce bugs if the classes are unrelated.
Find more details in virtual function specifier.
Related
I want to initialize a variable class for my Base class and modify it only in some children classes.
The initialization of this class variable is in the header:
class Base{
public:
Base();
int a = 1;
The header of my derived class is:
class ChildA : public Base{
public:
ChildA ();
int a = 2;
}
Problem
I tried to run this:
Base classe*;
classe = new ChildA();
std::cout << classe->a << std::endl;
The problem is that, instead of printing 2 as I expected, it prints 1 (value initialized in my parent class). For other derived classes, I want classe->a to still return 1.
How can I solve this?
What you are seeing are the results of upcasting - having a base class point to a derived class object.
I highly recommend you read more on the topic, but a simplified picture is that the base class "carries" the whole object (base and derived content), but can access only it's original, own content. This is why you see a value from the base class rather than the derived class.
Lippman's C++ Primer has a very comprehensive explanation of upcasting, downcasting, object slicing, and other inheritance-related concepts. This includes implementing virtual functions which give you the functionality to invoke derived class functions through an interface that's common with base.
|----------------|-----------------|
| a (for Base) | a (for ChildA) |
|----------------|-----------------|
\________________/
^
|_ classe
Sorry if my drawing is not that good! As you can see in the above picture, when you create an object of type ChildA, this object contains a part for keeping the data members of the base class (i.e. Base) and another part for the data members of derived class (i.e. ChildA). Considering the fact that you have a pointer to the base class (i.e. classe), this pointer only allows you to access the base member variables and methods. So calling classe->a, returns the Base::a for you, that is 1.
If you change the type of classe pointer into ChildA, in that case calling classe->a will return 2, because it refers to the a inside the derived class, i.e. ChildA::a.
For other derived classes, I want classe->a to still return 1. How can
I solve this?
If you need in some derived classes to access different as, it is better to have a virtual function in the base class and override the base function if needed:
class Base {
public:
Base() {}
virtual int getA() { return a; }
private:
int a = 1;
};
class ChildA : public Base
{
public:
ChildA() {}
int a = 2;
};
class ChildB : public Base
{
public:
ChildB() {}
int getA() { return a; }
private:
int a = 2;
};
Now by calling getA() on a pointer of Base or ChildA, you will have 1, but calling getA on an object of type ChildB will return 2.
This question already has answers here:
Should I initialize variable within constructor or outside constructor [duplicate]
(11 answers)
Closed 6 years ago.
What is the difference between below 2 ways of assigning values for variables of a class.
Class A{
Private Variable v = someValue;
}
vs
Class A{
private Variable v;
//constructor
public A(){
this.v = someValue;
}
}
Can someone please explain?
There is no real difference from a code execution point of view.
As a previous answer says, I prefer declaring the variable outside of the constructor; for example:
public class A {
private int aValue = 100;
}
Instead of
public class A {
private int aValue;
public A() {
this.aValue = 100;
}
}
The reason being that if you have multiple constructors, you do not have to keep writing this.aValue = 100; and you are unable to "forget" to initialize the variable in a constructor.
As others have said however, there are times when it is better to initialize the variable in the constructor.
If it will change based on values passed to it via the constructor, obviously initialize it there.
If the variable you are initializing may throw an error and you need to use try / catch - it is clearly better to initialize it in the constructor
If you are working on a team that uses a specific coding standard and they require you to initialize your variables in the constructor, you should do so.
Given freedom and none of the above, I still declare it at the top - makes it much easier to find all of your variables in one place (in my experience).
See this duplicate answer: https://stackoverflow.com/a/3919225/1274820
What is the difference between below 2 ways of assigning values for
variables of a class.
Generally nothing, but ...
class constructor is an entry point when creating a new instance, so all assignments should be done there for readability and maintainability.
When you want create a new instance you start reading a source code at the constructor. Here is an example. All informations about new instance are in one proper place.
public class C {
private int aValue;
private int bValue;
private int cValue;
private int dValue;
public C(int a, int b) {
this.aValue = a;
this.bValue = b;
this.cValue = a * b;
this.dValue = 1000;
}
}
If you look at the MSIL of this class:
namespace Demo
{
public class MyClass
{
private string str = "hello world";
private int b;
public MyClass(int b)
{
this.b = b;
}
}
}
.method public hidebysig specialname rtspecialname
instance void .ctor(int32 b) cil managed
{
// Code size 25 (0x19)
.maxstack 8
IL_0000: ldarg.0
IL_0001: ldstr "hello world"
IL_0006: stfld string Demo.MyClass::str <---- RIGHT HERE
IL_000b: ldarg.0
IL_000c: call instance void [mscorlib]System.Object::.ctor()
IL_0011: ldarg.0
IL_0012: ldarg.1
IL_0013: stfld int32 Demo.MyClass::b
IL_0018: ret
} // end of method MyClass::.ctor
You can see that the constructor is "injected" with the assignment of this.str = "hello world".
So once your code is compiled, there is no difference what so ever. Yet, there are quite a few good reasons why you should not do it (user1274820's answer has some them)
I have C++ code (not mine, so it is not editable). Problem is with extending protected functions and class.
#include "ExtraClass.h"
...
MyClass::MyClass()
{
...
protected:
bool Func{}
ExtraClass m_Foo;
...
}
I need access in Python to m_Foo methods and protected functions like Func() like
from MyClass import *
bar = MyClass()
bar.m_Foo.Run() //something like this
but have an compiler error:
*error: ‘ExtraClass MyApp::m_Foo’ is protected*
PS. If I change protected with public (just for try). I can access *m_Foo* only in readonly mode:
class_<MyClass>("MyClass", init<>())
.def_readonly("m_Foo", &MyClass::m_Foo)
Changing to *def_readwrite* went to compiler error:
/boost_1_52_0/boost/python/data_members.hpp:64:11: error: no match for ‘operator=’ in ‘(((ExtraClass)c) + ((sizetype)((const boost::python::detail::member<ExtraClass, MyClass>*)this)->boost::python::detail::member<ExtraClass, MyClass>::m_which)) = d’
Thank you for any help!
In general, if you want to wrap protected members, then you need to derive a (wrapper) class from the parent that makes the members public. (You can simply say using Base::ProtectedMember in a public section to expose it instead of wrapping it). You will then have wrap it normally. Like this:
class MyWrapperClass : public MyClass {
public:
using MyClass::m_Foo;
};
In this particular example (which is really not fully baked), if you want to access m_Foo, then you need to wrap ExtraClass. Assuming that you have The problem with readwrite is likely the implementation of ExtraClass (which probably doesn't supply a operator= that you can use).
I'm loading a .NET assembly dinamically via reflection and I'm getting all the classes that it contains (at the moment one). After this, I'm trying to cast the class to an interface that I'm 100% sure the class implements but I receive this exception: Unable to cast object of type System.RuntimeType to the type MyInterface
MyDLL.dll
public interface MyInterface
{
void MyMethod();
}
MyOtherDLL.dll
public class MyClass : MyInterface
{
public void MyMethod()
{
...
}
}
public class MyLoader
{
Assembly myAssembly = Assembly.LoadFile("MyDLL.dll");
IEnumerable<Type> types = extension.GetTypes().Where(x => x.IsClass);
foreach (Type type in types)
{
((MyInterface)type).MyMethod();
}
}
I have stripped out all the code that is not necessary. This is basically what I do. I saw in this question that Andi answered with a problem that seems the same mine but I cannot anyway fix it.
You are trying to cast a .NET framework object of type Type to an interface that you created. The Type object does not implement your interface, so it can't be cast. You should first create a specific instance of your object, such as through using an Activator like this:
// this goes inside your for loop
MyInterface myInterface = (MyInterface)Activator.CreateInstance(type, false);
myInterface.MyMethod();
The CreateInstance method has other overloades that may fit your needs.
I am having confusion with the following code:
class A
{
int x;
static void F(B b) {
b.x = 1; /* Ok,
I want to know how is this ok, in a static block how a non static
instance variables are called because as I know that static block
gets memory at compile time before execution of a single command
while non static at run time and static method accessing a non static
variable which is not created yet please elaborate me on this
*/
}
}
class B: A
{
static void F(B b) {
b.x = 1; // Error, x not accessible
}
}
Nothing gets memory at compile time. Static fields are indeed placed in the static block of memory when the type gets initialized. Call stacks for static methods are allocated at run time exactly like in case of instance methods.
Now, why static methods don't have access to the instance fields. Consider this:
class A {
public int Value;
static int GetValue() {
return Value;
}
}
There you have a class with an instance field and a static method. Now, somewhere else you try this:
var a1 = new A();
a1.Value = 5;
var a2 = new A();
a2.Value = 10;
int result = A.GetValue();
Now, if compiler allowed this, what value would the result get? 5 or 10 or something else? This just doesn't make sense, because static methods are declared for class as a whole and aren't aware of instances of this class. So in the code of static method you don't know how many (if any) instances of this class exist and can't access their instance fields.
Hope this makes a little sense.
Either you changed the code in question a bit or I didn't read very carefully. Seems like it's completely different problem right now. The variable x is indeed not accessible for the class B because of its level of protection (default in C# is private). Class A can modify X because it's declared in class A and visible to its method. Class B can't do it (you must make x protected or public for that).