Is it possible to edit template name at compile time? [closed] - c++

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Closed 10 years ago.
I am trying to modify template type given at compile time but unable to do it. Let see if you have some idea.
Lets consider we have 2 classes class A and A_test and one template class B.
Will implemented class B as it is having a function b_f() which is internaly creating an object of template class and calling public funcion A_f()/A_test_f() of class A or Class A_test.
from main, will be creating class B obj; obj.b_f();
but I want to creat an object of class A_test not class A.
Let me know if it is possible.
Basicaly I wanted to do object injection. Please let me know if it is possible.

The best solution by far is to try to rename A_test_f() to be the same as A_f().
If that proves impossible, the next thing I'd try is to specialize class B:
template<class AT>
class B {
public: b_f() {
AT m_A;
m_A.A_f();
};
template<>
class B<A_test> {
public: b_f() {
AT m_A;
m_A.A_test_f();
};
If B is too complicated, there are other things to try, but you probably want to rethink what you're doing first. If all else fails, do what I coded above here, but call it B_HELPER instead of B and then b_f() can do: B_HELPER<AT> m_A; m_A.b_f(); That way you don't have to recode all of B.

The common way to inject dependencies into templates at compile-time is through type traits. This allows customizing the templates externally through another struct or class which has specific knowledge of the class. Examples in the standard library include std::char_traits<> and std::iterator_traits<>. Boost also defines some, including boost::type_traits<>.
Traits involve defining a struct for the general case and specializing it for alternate cases when necessary.
// general case: select method named "f".
template<class T> struct b_traits
{
typedef void(T*F)();
static const F f = &T::f;
};
// template type that forwards method selection to "b_traits" struct.
template<class AT>
class B {
public: b_f() {
AT m_A;
(m_A.*(b_traits<AT>::f))();
};
class A_test { ... };
// special case: select method named "A_test_f".
template<> struct b_traits<A_test>
{
typedef void(T*F)();
static const F f = &A_test::A_test_f;
};
int main ()
{
B<A_test> b;
b.b_f(); // will invoke "A_test::A_test_f()" rather than "A_test::f()".
}

Related

How do I handle classes that reference each other? [closed]

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Closed 9 years ago.
Here's the code:
class B
{
A a;
};
class A
{
B b;
};
int main()
{
return 0;
}
Here's the error:
1>c:\mine\visual studio 2010 projects\myproj\compiled equivalent.cpp(7):
error C2079: 'B::a' uses undefined class 'A'
You can't. Two classes cannot contain each other as members. Consider answering the question “What is the size of the type A?” Well A contains a B, so what is the size of B? Well B contains an A, so what is the size of A? Oh dear, we have an infinite loop. How would we possibly store this object in finite memory?
Perhaps a more appropriate structure would be to have one of the classes contain a pointer to the other type. The type that is pointed to can simply be forward declared before the pointer member is declared:
class A; // A is only declared here, so it is an incomplete type
class B
{
A* a; // Here it is okay for A to be an incomplete type
};
class A
{
B b;
};
Now the type B doesn't contain an A, it just contains a pointer to an A. There doesn't even have to be an A object for it to point to, so we've broken the infinite cycle.
Considering you ask for references between the classes, probably you come from Java, C# or similar background where one can only put references to objects inside other objects.
In C++ there is no such restriction: you are allowed to have one object's content completely nested inside another one. But for this to work you must provide the definition of the nested object beforehand. C++ needs this definition in order to compute the size and layout of the outer object. In order to escape from this nesting you need to place not the object itself but a pointer or reference to it inside the outer object.
That being said,
// C#
class A
{
int x;
B b;
}
class B
{
int y;
A a;
}
becomes
// C++
class B; // tell the compiler that B is a class name
class A
{
int x;
B *pb; // the forward declaration above allows you to declare pointer to B here
};
class B
{
int y;
A *pa;
};
if you decide to use the pointer syntax.
What this allows is to have things like:
// C++ again
class C
{
A a;
B b;
A *pa2;
};
which has memory layout in the form of:
C:
a.xa.pb
b.yb.pa
pa2
which is not possible in Java/C#.

Implement several order relations for one class? [closed]

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Closed 10 years ago.
Which is the best way to implement several order relations for only one class? I have an idea with the Strategy Pattern but I'm not sure that's a good idea. And if there is not a best way, why?
Create a functor class and initialize a member with the order relation you want to use. Have operator() use the member to decide the ordering of the two operands.
class Ordering
{
public:
Ordering(int method) : _method(method) {}
bool operator()(const MyObject & first, const MoObject & second) const
{
switch(_method)
{
case 0:
return first.name < second.name;
case 1:
return first.age < second.age;
// ...
}
}
int _method; // an enum would be better
};
std::sort(myobjs.begin(), myobjs.end(), Ordering(selected_method));
I think Strategy is a better way here, and I'm not pretty sure that a switch structure is a good idea (imagine, 1000 comparison methods in one switch... Too heavy, isn't it?)
So let A, a class which need a method comparison.
I suggest to create one class per method, which instance will be A's component.
For instance :
class A{
private:
//Some attributes
Comparator<A> comp_;
public:
//Some methods (including constructor)
bool operator()(const MyObject & first, const MoObject & second) const
{
return comp_.compare(first,second);
}
void setComparator(Comparator<A>& comp){
comp_ = comp;
}
}
//Forgot the syntax about template. So there is a template with one parameter
class Comparator{
public:
//Constructor
//To overwrite on subclasses
virtual bool compare(T& first, T& second) = 0;
}
With that configuration, you can easily add a method, without modifying A, just set the right comparator at any moment of program's execution.

C++ polymorphism: going from base class to derived class [closed]

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Closed 10 years ago.
class A{};
class B : A{};
void func(A* p)
{
B* p2 = p; // Error
}
Your code has several oddities.
You use private inheritance. In private inheritance you will not be able to convert to a derived class ever when you are not inside the class scope itself.
Even if you would inherit publicly, you will need at least one virtual function (and that should be the destructor) in the base class to use dynamic_cast.
Chances are you are doing something wrong when you need a lot of down-casts. You should probably rethink your design or usage of the provided API.
Typically, things would look like this:
class A {
public:
virtual ~A() {}
};
class Derived : public A {
};
void func(A* a) {
if(Derived* d = dynamic_cast<Derived*>(a)) {
// yeah, a is of type derived
} else {
// a is not of type Derived
}
}
is-a relationship is implemented by public inheritance. as you are inheriting it privately this is association relationship which is not is-a. so B is not an A. So you cannot store A in B or A cannot become an B
and dynamic_cast will not work as source type is not polymorphic
neither would static_cast work as ‘A’ is an inaccessible base of ‘B’
To make an is-a relationship you need to do class B: public A{}
For classes private inheritance is used by default.

Right way to instantiate instance-variables in C++ [closed]

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Closed 11 years ago.
I want to ask where in C++ is the right place to instantiate a instance-variables? I think it should not be in the class declaration, but otherwise I don`t see any disadvantages apart from poor object-oriented design:
class A{ member m; };
I think it should better be like:
class A{ extern member m; };
But I don`t know how to realize it without a pointer like this:
class A{ member* m };
A::A(){ m = new member; }
Is there a "clean solution" to realize this on the stack (without using pointers)?
You can use the constructor initialization list to construct all your member variables as you need.
A::A(const member& memberArg)
: m(memberArg)
{ }
Look at this for more info.
I think you have a misunderstanding of how objects are instantiated. If all you do is declare a class, no member variables are actually instantiated. It isn't until you construct an instance of that class that its member variables exist.
Here's an example to show when a member object gets instantiated:
class ClassA
{
public:
ClassA() { std::cout << "Hello!\n"; }
};
class ClassB
{
public:
ClassA objA;
};
int main()
{
// do some work
ClassB objB; // here, a ClassB object is created, and with it its member ClassA object, so "Hello!" is printed
return 0;
}
As to exactly how you specify what kind of ClassA object to create if its constructor requires arguments, the other answers do a fine job explaining it.
I think it should not be in the class declaration, but otherwise I don`t see any disadvantages apart from poor object-oriented design:
class A{ member m; };
What in your mind makes this poor OO design? This is the preferred mechanism in C++.
I think it should better be like:
class A{ extern member m; };
This isn't valid code. Qualifying member data with a storage class specification such as extern is illegal.
But I don`t know how to realize it without a pointer like this:
class A{ member* m; };
A::A(){ m = new member; }
That will work, but why do that? It looks to me like you are trying to import a Java POV into C++. Everything is allocated, and everything is a reference in Java. In many (most!) cases there is no reason to allocate data members in C++. All it does is add an unneeded indirection and add a place where memory can leak.
You want to use member initializers: they are the only way to initialize class members that have a constructor that requires parameters, and the cleanest way to initialize other class members.
If you have class A and a member m with a constructor:
class A { member m; }
You want
class A { member m; A(); }
A::A()
: m(<constructor params>)
{
}
You would declare your instance variables in your .h file:
A.h
class A {
public:
A();
private:
int value;
double someOtherValue;
}
You can instantiate them in your .cpp file like so:
A.cpp
A::A(): value(5), someOtherValue(10.0)
{
...
}
If the member object is to be totally controlled by the enclosing A object, your first example is the proper way to do it. It does have the downside of requiring a complete definition of member at the point where A is defined.
You could check out the pimpl idiom to reduce coupling, but that still requires the object to be heap-based and not stack-based.

c++ access stuff out side the class [closed]

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Closed 12 years ago.
ok, lets say I have this:
class A:
{
Public:
int dd;
}
Ok, then I have this class:
class B:
{
Public:
void sss(){ ff=dd; }
int ff;
}
Ok, the problem is that class B doesn't know what dd is. How can I make class B know it?
It's public: and not Public:.
Because dd is part of A, you need an instance of A to access dd.
The two classes aren't related in any way; since B does not have access to any A objects it can't take the value of its members.
If you were to pass an A object to the B constructor, either by value, reference, or pointer, you could access its dd member since you made it public.
You need to have an instance of A, and pass it to the sss method in order to access this data:
void sss(A a) { ff = a.dd; }
If you want to have only one copy of dd, rather than a single copy per instance of A, then you'll have to make dd static:
class A
{
public:
static int dd;
};
class B
{
public:
void sss() { ff = A::dd; }
int ff;
};
Maybe by passing in an A instance?
class B:
{
Public:
void sss(const A& a){ ff=a.dd; }
int ff;
}
Your code does not look like C++ but I will try to answer your question.
There would be several possibilities to do what you want. I just pick one which is easy to understand.
Do someting like this:
class B
{
public:
void sss() { ff = a.dd; }
int ff;
private:
A a;
};
However if you do not tell me what you really want to achieve with that class i.e. which responsibilites the class has this will not help you much.
The code that you present, ...
class A:
{
Public:
int dd;
}
class B:
{
Public:
void sss(){ ff=dd; }
int ff;
}
... is not valid C++.
You can find that out easily enough by trying to compile it. Then the compiler will complain about a lot of irrelevant things, but its first message will be related to the extraneous colon after A. When you remove that and compile again, its first message will be related to Public, as opposed to correct public (lowercase). So on.
Syntacically correct code (that will still produce a compilation error, but not about the basic syntax):
class A
{
public:
int dd;
};
class B
{
public:
void sss(){ ff=dd; }
int ff;
};
Imagine that you create five instances of class A, named a1, a2, a3, a4, and a5. Each such instance has a member dd, so you can refer to these five members as a1.dd, a2.dd, a3.dd, a4.dd, and a5.dd.
Now you create a B instance, naming it 'b'. And you call b.sss(). And that member function effectively does b.ff=dd. Or it would, had it been meaningful and accepted by the compiler. But it can't, for which of a1.dd, a2.dd, a3.dd, a4.dd, and a5.dd is being referred to?
So, since this is a very basic concept in C++, you need a good C++ textbook, such as Bjarne's latest, and start from the beginning.
Alternatively, since C++ is a very complicated programming language, it's probably even better to start with some simpler language, like Java, or even simpler, like JavaScript or Python. I recommend Python. It's very different from C++ in the details, but both languages are "conventional" programming languages. Using Python you'll learn about decision constructs and functions and things, including classes versus instances.
Cheers & hth.,