Using boost::function with a parameter to shared pointer to derived class

问题

Using C++ with g++ 5.4.0 on Ubuntu 16.04.

I have a class A, and a class B that derives from class A. Function f1 takes a shared pointer to class A as parameter. Function f2 takes a shared pointer to class B as parameter and return the same type as f1. Using boost::function, another function F takes a function such as f1 as parameter. The code looks like this :

result_t f1 ( const boost::shared_ptr<A> a );
result_t f2 ( const boost::shared_ptr<B> b );
typedef boost::function < result_t (const boost::shared_ptr<A>&) > f1_t;
void F ( const f1_t f );

Calling F with f1 as argument works fine. I now want to call F but with f2 as argument. I get the following error :

error: invalid initialization of reference of type const boost::shared_ptr<B>& from expression of type const boost::shared_ptr<A>
           return f(BOOST_FUNCTION_ARGS);

There's not really a need for F to get this error, doing :

f1_t f = f2;

gives the same error.

回答1:

There's no co-variance for function prototypes. Different signatures are simply that: different types.

In this case you'd need to wrap the function with a converting wrapper.

Let's create a few facility definitions:

using result_t = int;
struct A { };
struct B : A { };

typedef boost::shared_ptr<A> APtr;
typedef boost::shared_ptr<B> BPtr;

result_t f1(APtr) { return 1; }
result_t f2(BPtr) { return 2; }

typedef boost::function <result_t(APtr const&)> funOfA;
typedef boost::function <result_t(BPtr const&)> funOfB;

Now wrapping funOfB as a funOfA would look like this:

funOfA wrapFunOfB(const funOfB f) {
    struct {
        funOfB _f;
        result_t operator()(APtr const& a) const { 
            return _f(boost::static_pointer_cast<B>(a));
        }
    } wrap { f };

    return wrap;
}

Now you can easily write:

int main() {
    F(f1);
    F(wrapFunOfB(f2));
}

Simple C++03 Demo

Live On Coliru

#include <boost/shared_ptr.hpp>
#include <boost/make_shared.hpp>
#include <boost/function.hpp>
#include <iostream>

typedef int result_t;
struct A { int i; };
struct B : A { int j; };

typedef boost::shared_ptr<A> APtr;
typedef boost::shared_ptr<B> BPtr;

result_t f1(APtr) { return 1; }
result_t f2(BPtr) { return 2; }

typedef boost::function <result_t(APtr const&)> funOfA;
typedef boost::function <result_t(BPtr const&)> funOfB;

struct Wrapper {
    typedef result_t result_type;
    funOfB _f;

    result_t operator()(APtr const& a) { 
        return _f(boost::static_pointer_cast<B>(a));
    }
};

funOfA wrapFunOfB(const funOfB f) {
    Wrapper wrap = { f };
    return wrap;
}

void F(const funOfA f) {
    APtr a = boost::make_shared<A>();
    APtr b = boost::make_shared<B>();

    //std::cout << "f(a): " << f(a) << "\n"; // UNDEFINED BEHAVIOUR if f wraps a funOfB
    std::cout << "f(b): " << f(b) << "\n";
}

int main() {
    F(f1);
    F(wrapFunOfB(f2));
}

Prints

f(b): 1
f(b): 2

PROBLEMS, WARNINGS: dynamic_pointer_cast<>

If F actually invokes the parameter on an object that isn't actually of type B, that static_cast<> will invoke Undefined Behaviour.

If you want to protect against that, use dynamic_pointer_cast, which requires the classes A and B to be polymorphic types.

Live On Coliru

#include <boost/shared_ptr.hpp>
#include <boost/make_shared.hpp>
#include <boost/function.hpp>
#include <iostream>

typedef int result_t;
struct A     { int i; virtual ~A() {} };
struct B : A { int j; };

typedef boost::shared_ptr<A> APtr;
typedef boost::shared_ptr<B> BPtr;

result_t f1(APtr a) { return a?1 : 0; }
result_t f2(BPtr b) { return b?2 : -99; }

typedef boost::function <result_t(APtr const&)> funOfA;
typedef boost::function <result_t(BPtr const&)> funOfB;

struct Wrapper {
    typedef result_t result_type;
    funOfB _f;

    result_t operator()(APtr const& a) { 
        return _f(boost::dynamic_pointer_cast<B>(a));
    }
};

funOfA wrapFunOfB(const funOfB f) {
    Wrapper wrap = { f };
    return wrap;
}

void F(const funOfA f) {
    APtr a = boost::make_shared<A>();
    APtr b = boost::make_shared<B>();

    std::cout << "f(a): " << f(a) << "\n";
    std::cout << "f(b): " << f(b) << "\n";
}

int main() {
    F(f1);
    F(wrapFunOfB(f2));
}

Prints

f(a): 1
f(b): 1
f(a): -99
f(b): 2

C++11 Version

Things get a little more elegant here. Notably, the Wrapper class can be local, and anonymous:

funOfA wrapFunOfB(const funOfB f) {
    struct {
        typedef result_t result_type;
        funOfB _f;

        result_t operator()(APtr const& a) { 
            return _f(std::dynamic_pointer_cast<B>(a));
        }
    } wrap { f };
    return wrap;
}

Next level: use a lambda instead:

funOfA wrapFunOfB(const funOfB f) {
    return [f](APtr const& a) { return f(std::dynamic_pointer_cast<B>(a)); };
}

Live On Coliru

#include <memory>
#include <functional>
#include <iostream>

typedef int result_t;
struct A     { int i; virtual ~A() {} };
struct B : A { int j; };

typedef std::shared_ptr<A> APtr;
typedef std::shared_ptr<B> BPtr;

result_t f1(APtr a) { return a?1 : 0; }
result_t f2(BPtr b) { return b?2 : -99; }

typedef std::function<result_t(APtr const&)> funOfA;
typedef std::function<result_t(BPtr const&)> funOfB;

funOfA wrapFunOfB(const funOfB f) {
    return [f](APtr const& a) { return f(std::dynamic_pointer_cast<B>(a)); };
}

void F(const funOfA f) {
    APtr a = std::make_shared<A>();
    APtr b = std::make_shared<B>();

    std::cout << "f(a): " << f(a) << "\n";
    std::cout << "f(b): " << f(b) << "\n";
}

int main() {
    F(f1);
    F(wrapFunOfB(f2));
}

That's 25% code reduction.

来源：https://stackoverflow.com/questions/49325968/using-boostfunction-with-a-parameter-to-shared-pointer-to-derived-class