How to avoid unnecessary construction of std::vector<Base *> - c++

I have two classes
class Base{ ... };
class Derived : public Base { ... };
and some other function which operates and modifies a collection of Base pointers:
void foo(std::vector<Base *> ptrs_to_base);
I have a std::vector<Derived> which holds my data. To pass it to foo I can construct a new vector of type std::vector<Base *> containing all pointers. In many situations this will be the easiest thing to do. However, I don't like the overhead with which it comes and in some situations it is simply not affordable to allocate dynamic memory.
Is there a way to change foo's interface as to do everything in place? In principle there should be no need for unnecessary construction of collections. I know there is no easy way, so usage of additional (multi-purpose) helper classes would be fine.
For example
template <typename Iter>
void foo(Iter begin, Iter end) {
//*begin is of type Base*
...
}
std::vector<Derived> elements;
std::vector<Derived *> ptrs_to_elements = ... // construct from elements
foo(ptrs_to_elements.begin(), ptrs_to_elements.end());
would be one solution which requires templates and doesn't make it obvious that only pointers of type Base * are required. Edit: W.F. gave a way to recognize wrong usages of this template. It still requires the unnecessary construction of ptrs_to_elements.

The problem is that your function takes a vector as parameter. And vector as any other standard container does not implement a simple interface and is not intended to be derived.
That means that if you only have a vector of objects and need a vector of pointers your only choice is to build a new vector.
For that reason, standard C++ container are very good to store objects, but public interface should rather use... interfaces of simple wrappers around them to make on flight transformation easier (if not simply just possible)

Related

How to pass a container of derived type to a function needing a container of base type [duplicate]

Consider these classes.
class Base
{
...
};
class Derived : public Base
{
...
};
this function
void BaseFoo( std::vector<Base*>vec )
{
...
}
And finally my vector
std::vector<Derived*>derived;
I want to pass derived to function BaseFoo, but the compiler doesn't let me. How do I solve this, without copying the whole vector to a std::vector<Base*>?
vector<Base*> and vector<Derived*> are unrelated types, so you can't do this. This is explained in the C++ FAQ here.
You need to change your variable from a vector<Derived*> to a vector<Base*> and insert Derived objects into it.
Also, to avoid copying the vector unnecessarily, you should pass it by const-reference, not by value:
void BaseFoo( const std::vector<Base*>& vec )
{
...
}
Finally, to avoid memory leaks, and make your code exception-safe, consider using a container designed to handle heap-allocated objects, e.g:
#include <boost/ptr_container/ptr_vector.hpp>
boost::ptr_vector<Base> vec;
Alternatively, change the vector to hold a smart pointer instead of using raw pointers:
#include <memory>
std::vector< std::shared_ptr<Base*> > vec;
or
#include <boost/shared_ptr.hpp>
std::vector< boost::shared_ptr<Base*> > vec;
In each case, you would need to modify your BaseFoo function accordingly.
Instead of passing the container object (vector<>), pass in begin and end iterators like the rest of the STL algorithms. The function that receives them will be templated, and it won't matter if you pass in Derived* or Base*.
This problem occurs in programming languages that have mutable containers. You cannot pass around a mutable bag of apples as a bag of fruit because you cannot be sure that someone else does not put a lemon into that bag of fruit, after which it no longer qualifies as a bag of apples. If the bag of apples were not mutable, passing it around as a bag of fruit would be fine. Search for covariance/contravariance.
one option is to use a template
template<typename T>
void BaseFoo( const std::vector<T*>& vec)
{
...
}
The drawback is that the implementation has to be in the header and you will get a little code bloat. You will wind up with different functions being instantiated for each type, but the code stays the same. Depending on the use case it's a quick and dirty solution.
Edit, I should note the reason we need a template here is because we are trying to write the same code for unrelated types as noted by several other posters. Templates allow you do solve these exact problems. I also updated it to use a const reference. You should also pass "heavy" objects like a vector by const reference when you don't need a copy, which is basically always.
Generally you would start with a container of base pointers, not the other way.
If you dealing with a third-party library, and this is your only hope, then you can do this:
BaseFoo (*reinterpret_cast<std::vector<Base *> *>(&derived));
Otherwise fix your code with one of the other suggesstions.
Taking Matt Price's answer from above, given that you know in advance what types you want to use with your function, you can declare the function template in the header file, and then add explicit instantiations for those types:
// BaseFoo.h
template<typename T>
void BaseFoo( const std::vector<T*>& vec);
// BaseFoo.cpp
template<typename T>
void BaseFoo( const std::vector<T*>& vec);
{
...
}
// Explicit instantiation means no need for definition in the header file.
template void BaseFoo<Base> ( const std::vector<Base*>& vec );
template void BaseFoo<Derived> ( const std::vector<Derived*>& vec );
If std::vector supported what you're asking for, then it would be possible to defeat the C++ type system without using any casts (edit: ChrisN's link to the C++ FAQ Lite talks about the same issue):
class Base {};
class Derived1 : public Base {};
class Derived2 : public Base {};
void pushStuff(std::vector<Base*>& vec) {
vec.push_back(new Derived2);
vec.push_back(new Base);
}
...
std::vector<Derived1*> vec;
pushStuff(vec); // Not legal
// Now vec contains a Derived2 and a Base!
Since your BaseFoo() function takes the vector by value, it cannot modify the original vector that you passed in, so what I wrote would not be possible. But if it takes a non-const reference and you use reinterpret_cast<std::vector<Base*>&>() to pass your std::vector<Derived*>, you might not get the result that you want, and your program might crash.
Java arrays support covariant subtyping, and this requires Java to do a runtime type check every time you store a value in an array. This too is undesirable.
They are unrelated types -- you can't.

Modifying member functions of built-in objects

I'm sure this has been answered many times, but I don't know how to search for this. Is this polymorphism? Overloading?
Basically I want to take a vector, and modify its functions the way I want them or create new ones.
E.g: I want push_back() to insert an element and keep the vector in descending order. So the addition would be to swap() the element to its respective place. Or I would like to add a new function pop_front().
That being said, I need it as a member function, e.g:
vector<int> x;
x.pop_front();
not:
pop_front(x);
Is this ever done in practice? I know I can just use existing containers, like a priority queue for my example, but I'd rather make completely custom functions.
Basically, std::vector is not designed to be base class, inherit from it would be a bad idea in general. For example its dtor is not virtual.
It's better to use composition here. e.g.
template <typename T>
class MyVector {
private:
std::vector<T> v;
public:
void pop_front() {
// processing on v ...
}
};

Using a(n empty) base class to enable storing different objects in the same container

Say I have two different objects that are completely different, Sprite and PhysicsData.
I write an empty base class and a container class that can push and remove objects to/from a container.
I create two of these containers to store the two different objects - Sprite and PhysicsData.
(Different objects aren't together in the same class)
class base
{
};
class ContainerManager
{
public:
std::vector<base*> list;
void Push(base *object);
void Remove(base *object);
};
class PhysicsData : public base
{
void applyGravity();
};
class Sprite : public base
{
void Draw();
};
ContainerManager SpriteContainer;
ContainerManager PhysicsDataContainer;
Sprite aSprite;
SpriteContainer.Push(&aSprite);
PhysicsData SomeData;
PhysicsDataContainer.Push(&SomeData);
Is this the way that this should bet done?
This is not a C++ way. You should use templates.
For you to know, STL (which you are calling to when using std:: namespace prefix) is, actually, Standard Template Library :). A lot of template classes are already there, e.g. for the push / remove operations see std::list<T>
You use it like this:
std::list<Sprite> SpriteContainer;
std::list<PhysicsData> PhysicsDataContainer;
and so on.
There is a cool guide about C++ templates, if you still want to do your own class for some more functionality.
And there is a reference to std::list, (i don't think i need to explain the usage of std::vector for you) if the question was the actual thing you've wanted to do.
You're having Templates in C++ and still worrying about having a common base class for a trivial container??
template <class T>
class Container{
private:
vector<T> list;
public:
void Push(T data);
T Pop();
};
If you put a single virtual function into the base class, you'll be able to use dynamic_cast to get back the proper pointer if you mix different types in the same container. A virtual destructor would be a good idea because then you could delete the object if it were dynamically allocated, without having to cast it back to the original pointer.
If you're not going to mix types within a single container, I agree with dreamzor that a template class would be better. That's the way all the standard containers are implemented.
That looks like technically correct code but you are not doing any type checking. So as you can see PhysicsData is showing up in your SpriteContainer. You probably do not want this.
There is more than one way to keep PhysicsData out of your SpriteContainer. One way is to use templates. With templates you would state what type of base objects the container should work with at compile time.
Another method is to inherit from your ContainerManager two types of ContainerManagers, one called PhysicsContainerManager and one called SpriteContainerManager. These two child classes can do type check to verify that the objects being passed are in fact either a SpriteContainer or a PhisicsData. This link shows how to do type check in c++ C++ equivalent of instanceof
thanks
jose

container of unrelated T in c++

If I have the following hypothetical class:
namespace System
{
template <class T>
class Container
{
public:
Container() { }
~Container() { }
}
}
If I instantiate two Containers with different T's, say:
Container<int> a;
Container<string> b;
I would like to create vector with pointers to a and b. Since a and b are different types, normally this wouldn't be possible. However, if I did something like:
std::stack<void*> _collection;
void *p = reinterpret_cast<void*>(&a);
void *q = reinterpret_cast<void*>(&b);
_collection.push(a);
_collection.push(b);
Then later on, I can get a and b back from _collection like so:
Container<string> b = *reinterpret_cast<Container<string>*>(_collection.pop());
Container<int> a = *reinterpret_cast<Container<int>*>(_collection.pop());
My question is, is this the best way for storing a collection of unrelated types? Also would this be the preferred way of storing and retrieving the pointers from the vector (the reinterpret cast)? I've looked around and seen that boost has a nicer way of solving this, Boost::Any, but since this is a learning project I am on I would like to do it myself (Also I have been curious to find a good reason to use a reinterpret_cast correctly).
Consider boost::any or boost::variant if you want to store objects of heterogeneous types.
And before deciding which one to use, have a look at the comparison:
Boost.Variant vs. Boost.Any
Hopefully, it will help you to make the correct decision. Choose one, and any of the container from the standard library to store the objects, std::stack<boost::any>, std::stack<boost::variant>, or any other. Don't write your own container.
I repeat don't write your own container. Use containers from the standard library. They're well-tested.
While it is possible to cast to void * and back, the problem is knowing which type you're popping. After all, you give the example:
Container<string> b = *reinterpret_cast<Container<string>*>(_collection.pop());
Container<int> a = *reinterpret_cast<Container<int>*>(_collection.pop());
However, if you were to accidentally do:
Container<int> a = *reinterpret_cast<Container<int>*>(_collection.pop());
Container<string> b = *reinterpret_cast<Container<string>*>(_collection.pop());
Now you've got pointers to the wrong type, and will likely see crashes - or worse.
If you want to do something like this, at least use dynamic_cast to check that you have the right types. With dynamic_cast, you can have C++ check, at runtime (using RTTI), that your cast is safe, as long as the types being casted (both before and after) have a common base type with at least one virtual method.
So, first create a common base type with a virtual destructor:
class ContainerBase {
public:
virtual ~ContainerBase() { }
};
Make your containers derive from it:
template <typename T>
class Container : public ContainerBase {
// ...
}
Now use a std::stack<ContainerBase *>. When you retrieve items from the stack, use dynamic_cast<Container<int> >(stack.pop()) or dynamic_cast<Container<string> >(stack.pop()); if you have the types wrong, these will check, and will return NULL.
That said, heterogeneous containers are almost always the wrong thing to be using; at some level you need to know what's in the container so you can actually use it. What are you actually trying to accomplish by creating a container like this?

Getting a vector<Derived*> into a function that expects a vector<Base*>

Consider these classes.
class Base
{
...
};
class Derived : public Base
{
...
};
this function
void BaseFoo( std::vector<Base*>vec )
{
...
}
And finally my vector
std::vector<Derived*>derived;
I want to pass derived to function BaseFoo, but the compiler doesn't let me. How do I solve this, without copying the whole vector to a std::vector<Base*>?
vector<Base*> and vector<Derived*> are unrelated types, so you can't do this. This is explained in the C++ FAQ here.
You need to change your variable from a vector<Derived*> to a vector<Base*> and insert Derived objects into it.
Also, to avoid copying the vector unnecessarily, you should pass it by const-reference, not by value:
void BaseFoo( const std::vector<Base*>& vec )
{
...
}
Finally, to avoid memory leaks, and make your code exception-safe, consider using a container designed to handle heap-allocated objects, e.g:
#include <boost/ptr_container/ptr_vector.hpp>
boost::ptr_vector<Base> vec;
Alternatively, change the vector to hold a smart pointer instead of using raw pointers:
#include <memory>
std::vector< std::shared_ptr<Base*> > vec;
or
#include <boost/shared_ptr.hpp>
std::vector< boost::shared_ptr<Base*> > vec;
In each case, you would need to modify your BaseFoo function accordingly.
Instead of passing the container object (vector<>), pass in begin and end iterators like the rest of the STL algorithms. The function that receives them will be templated, and it won't matter if you pass in Derived* or Base*.
This problem occurs in programming languages that have mutable containers. You cannot pass around a mutable bag of apples as a bag of fruit because you cannot be sure that someone else does not put a lemon into that bag of fruit, after which it no longer qualifies as a bag of apples. If the bag of apples were not mutable, passing it around as a bag of fruit would be fine. Search for covariance/contravariance.
one option is to use a template
template<typename T>
void BaseFoo( const std::vector<T*>& vec)
{
...
}
The drawback is that the implementation has to be in the header and you will get a little code bloat. You will wind up with different functions being instantiated for each type, but the code stays the same. Depending on the use case it's a quick and dirty solution.
Edit, I should note the reason we need a template here is because we are trying to write the same code for unrelated types as noted by several other posters. Templates allow you do solve these exact problems. I also updated it to use a const reference. You should also pass "heavy" objects like a vector by const reference when you don't need a copy, which is basically always.
Generally you would start with a container of base pointers, not the other way.
If you dealing with a third-party library, and this is your only hope, then you can do this:
BaseFoo (*reinterpret_cast<std::vector<Base *> *>(&derived));
Otherwise fix your code with one of the other suggesstions.
Taking Matt Price's answer from above, given that you know in advance what types you want to use with your function, you can declare the function template in the header file, and then add explicit instantiations for those types:
// BaseFoo.h
template<typename T>
void BaseFoo( const std::vector<T*>& vec);
// BaseFoo.cpp
template<typename T>
void BaseFoo( const std::vector<T*>& vec);
{
...
}
// Explicit instantiation means no need for definition in the header file.
template void BaseFoo<Base> ( const std::vector<Base*>& vec );
template void BaseFoo<Derived> ( const std::vector<Derived*>& vec );
If std::vector supported what you're asking for, then it would be possible to defeat the C++ type system without using any casts (edit: ChrisN's link to the C++ FAQ Lite talks about the same issue):
class Base {};
class Derived1 : public Base {};
class Derived2 : public Base {};
void pushStuff(std::vector<Base*>& vec) {
vec.push_back(new Derived2);
vec.push_back(new Base);
}
...
std::vector<Derived1*> vec;
pushStuff(vec); // Not legal
// Now vec contains a Derived2 and a Base!
Since your BaseFoo() function takes the vector by value, it cannot modify the original vector that you passed in, so what I wrote would not be possible. But if it takes a non-const reference and you use reinterpret_cast<std::vector<Base*>&>() to pass your std::vector<Derived*>, you might not get the result that you want, and your program might crash.
Java arrays support covariant subtyping, and this requires Java to do a runtime type check every time you store a value in an array. This too is undesirable.
They are unrelated types -- you can't.