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Conversion from STL vector of subclass to vector of base class

I am wondering if it is possible to convert a vector of derived class values to a vector of base class values. Specifically I want to be able to pass a vector of base class objects to a function whose formal parameters takes a vector of base class. It does not appear to be possible directly as the following code example produces an error (using g++):

#include <vector>

class A {
};

class B : public A {
};


void function(std::vector<A> objs) {
}

int main(int argc, char **argv) {
    std::vector<B> objs_b;
    objs_b.push_back(B());
    function(objs_b);
}

test.cc:16: error: conversion from ‘std::vector<B, std::allocator<B> >’ to non-scalar type ‘std::vector<A, std::allocator<A> >’ requested

I would like to be able to be able to call function without having to define a new vector with elements of type A, inserting my elements of type B or changing to a vector of pointers.

like image 648
Stephen L Avatar asked Apr 18 '11 02:04

Stephen L


2 Answers

No, it is not. vector<B> is not derived from vector<A>, regardless of the fact that B is derived from A. You will have to change your function somehow.

A more idiomatically C++ approach might be to template it and have it take a pair of iterators - this is why the various standard library functions (<algorithm> etc) work that way, because it separates the implementation of the algorithm from the kind of thing it's operating on.

like image 110
Peter Avatar answered Sep 18 '22 17:09

Peter


Sometimes there are ways around this. Take a loot at Boost and some of the template meta-programming packages they have or use. Specifically I'd look at is_base_of for these kinds of purposes combined with partial template specialization.

For instance, if you wish to do some template magic trickery:

template<typename T, bool Allowed>
struct TemplateVectorCode;

You make a forward declaration of a templated class. Then you make a specialization with a "true" boolean value:

template<typename T> struct TemplateVectorCode<T, true>{
    void operator(const std::vector<T>& myVector) const{ //definition in here }
};

Finally you use that with is_base_of to only instantiate instances of your template functor in the following manner:

template<typename T, typename Base>
struct MyFunction : TemplateVectorCode<T, boost::is_base_of<Base,T>::value>{
};

The important point to note is that since we have not defined a TemplateVectorCode<T, false> the compiler will throw a compile error if you attempt to use your functor with a class type T that does not derive from the type V. That means you can work with the same code in one place without having to write multiple versions of it.

(if I screwed up the partial template specialization, someone please edit it. I need to go to bed.)

like image 36
wheaties Avatar answered Sep 18 '22 17:09

wheaties