我可以合法地将成员函数指针转换为函数指针吗？

Question

I've inherited some C++ code and I've been tasked with getting rid of warnings.

Here we have a member function pointer being cast to a function pointer. I understand that member function pointers are "different" from function pointers, in that there is an implicit 'this' parameter involved under the hood. However my predecessor appears to have made explicit use of this fact, by casting from a member function pointer to a function pointer with an additional first parameter inserted.

My Questions are:

A) Can I get rid of the compiler warning?

B) To what extent is this code guaranteed to work?

I've cut it down to a small main.cpp for the purposes of this question:

#define GENERIC_FUNC_TYPE   void(*)(void)
#define FUNC_TYPE       int(*)(void *)

class MyClass
{
public:
    MyClass(int a) : memberA(a) {}
    int myMemberFunc()
    {
        return memberA;
    }

private:
    int memberA;
};

int main(int argc, char*argv[])
{
    int (MyClass::* memberFunc) () = &MyClass::myMemberFunc;
    MyClass myObject(1);
    std::cout << (myObject.*memberFunc)() << std::endl;
    // All good so far

    // Now get naughty, store it away in a very basic fn ptr
    void(*myStoredFunction)(void) = (GENERIC_FUNC_TYPE)memberFunc;  // Compiler warning

    // Reinterpret the fn pointer as a pointer to fn, with an extra object parameter
    int (*myExtractedFunction)(void*) = (FUNC_TYPE)myStoredFunction;

    // Call it
    std::cout << myExtractedFunction(&myObject) << std::endl;
}

The code compiles with one warning under g++, and as intended outputs two 1's:

main.cpp: In function ‘int main(int, char**)’:
main.cpp:27:53: warning: converting from ‘int (MyClass::*)()’ to ‘void (*)()’ [-Wpmf-conversions]
  void(*myStoredFunction)(void) = (GENERIC_FUNC_TYPE)memberFunc; // Compiler warning
                                                     ^

IMHO this code is making assumptions about the underlying mechanisms of the compiler. Or maybe these assumptions are valid for all C++ compilers - Can anyone help?

(In the actual code we're storing a whole bunch of function pointers by name in a map. These functions all have different signatures, which is why they are all cast to the same signature void(*)(void). This is analogous to the myStoredFunction above. They are then cast to the individual signatures at the point of calling, analogous to myExtractedFunction above.)

Answer 1

How about create functions which avoid the cast entirely:

template <typename C, void (C::*M)()>
void AsFunc(void* p)
{
    (static_cast<C*>(p)->*M)();
}

then

void(*myStoredFunction)(void) = &AsFunc<MyClass, &MyClass::myMemberFunc>;

In C++17, with some traits, you might even have template <auto *M> void AsFunc(void* p) and void(*myStoredFunction)(void) = &AsFunc<&MyClass::myMemberFunc>;

Answer 2

To answer the question in the title, no, you can't legally cast a pointer-to-member-function to a pointer-to-function. Presumably, that's what the "Compiler warning" on the line with that cast said.

A conforming compiler is required to issue a diagnostic when confronted with ill-formed code (that's a bit oversimplified), and this one did. It gave a warning. Having done that, the compiler is free to do something implementation-specific, which it seems to have done: it compiled the code into something that does what you were hoping for.

Compilers are free to represent pointers to member functions in any way that works, and for non-virtual functions, that could be just a "normal" pointer to function. But try that with a virtual function; I'll bet the consequences are more harsh.

Answer 3

Since you apparently need to call a function by name on some "untyped" object ( void* ) while passing in a number of arguments that differ by function, you need some kind of multiple-dispatch. A possible solution is:

#include <string>
#include <iostream>
#include <stdexcept>
#include <functional>
#include <utility>
#include <map>

template <typename Subj>
using FunctionMap = std::map<std::string, std::function<void (Subj&, const std::string&)>>;

class AbstractBaseSubject {
    public:
        virtual void invoke (const std::string& fName, const std::string& arg) = 0;
};

template <typename Class>
class BaseSubject : public AbstractBaseSubject {
    public:
        virtual void invoke (const std::string& fName, const std::string& arg) {
            const FunctionMap<Class>& m = Class::functionMap;

            auto iter = m.find (fName);
            if (iter == m.end ())
                throw std::invalid_argument ("Unknown function \"" + fName + "\"");

            iter->second (*static_cast<Class*> (this), arg);
        }
};

class Cat : public BaseSubject<Cat> {
    public:
        Cat (const std::string& name) : name(name) {}
        void meow (const std::string& arg) {
            std::cout << "Cat(" << name << "): meow (" << arg << ")\n";
        }

        static const FunctionMap<Cat> functionMap;
    private:
        std::string name;
};

const FunctionMap<Cat> Cat::functionMap = {
    { "meow", [] (Cat& cat, const std::string& arg) { cat.meow (arg);  } }
};

class Dog : public BaseSubject<Dog> {
    public:
        Dog (int age) : age(age) {}
        void bark (float arg) {
            std::cout << "Dog(" << age << "): bark (" << arg << ")\n";
        }

        static const FunctionMap<Dog> functionMap;
    private:
        int age;
};

const FunctionMap<Dog> Dog::functionMap = {
    { "bark", [] (Dog& dog, const std::string& arg) { dog.bark (std::stof (arg));  }}
};

int main () {
    Cat cat ("Mr. Snuggles");
    Dog dog (7);

    AbstractBaseSubject& abstractDog = dog;     // Just to demonstrate that the calls work from the base class.
    AbstractBaseSubject& abstractCat = cat;

    abstractCat.invoke ("meow", "Please feed me");
    abstractDog.invoke ("bark", "3.14");

    try {
        abstractCat.invoke ("bark", "3.14");
    } catch (const std::invalid_argument& ex) {
        std::cerr << ex.what () << std::endl;
    }
    try {
        abstractCat.invoke ("quack", "3.14");
    } catch (const std::invalid_argument& ex) {
        std::cerr << ex.what () << std::endl;
    }
    try {
        abstractDog.invoke ("bark", "This is not a number");
    } catch (const std::invalid_argument& ex) {
        std::cerr << ex.what () << std::endl;
    }
}

Here, all classes with functions to be called this way need to derive from BaseSubject (which is a CRTP ). These classes (here: Cat and Dog , let's call them "subjects") have different functions with different arguments ( bark and meow - of course more than one function per subject is possible). Each subject has its own map of string-to-function. These functions are not function pointers, but std::function<void (SubjectType&,const std::string&)> instances. Each of those should call the respective member function of the object, passing in the needed arguments. The arguments need to come from some kind of generic data representation - here, I chose a simple std::string . It might be a JSON or XML object depending on where your data comes from. The std::function instances need to deserialize the data and pass it as arguments. The map is created as a static variable in each subject class, where the std::function instances are populated with lambdas. The BaseSubject class looks up the function instance and calls it. Since the subject class should always directly derive from BaseSubject<Subject> , pointers of type BaseSubject<Subject>* may be directly and safely cast to Subject* .

Note that there is no unsafe cast at all - it is all handled by virtual functions. Therefore, this should be perfectly portable. Having one map per subject class is typing-intensive, but allows you to have identically-named functions in different classes. Since some kind of data-unpacking for each function individually is necessary anyways, we have individual unpacking-lambdas inside the map .

If a function's arguments are just the abstract data structure, ie const std::string& , we could leave the lambdas out and just do:

const FunctionMap<Cat> Cat::functionMap = {
    { "meow", &Cat::meow }
};

Which works by way of std::function s magic (passing this via the 1st argument), which, in contrast to function pointers, is well-defined and allowed. This would be particularly useful if all functions have the same signature. In fact, we could then even leave out the std::function and plug in Jarod42's suggestion.

PS: Just for fun, here's an example where casting a member-function-pointer to an function-pointer fails:

#include <iostream>

struct A {
    char x;
    A () : x('A') {}
    void foo () {
        std::cout << "A::foo() x=" << x << std::endl;
    }
};

struct B {
    char x;
    B () : x('B') {}
    void foo () {
        std::cout << "B::foo() x=" << x << std::endl;
    }
};

struct X : A, B {
};

int main () {
    void (B::*memPtr) () = &B::foo;
    void (*funPtr) (X*) = reinterpret_cast<void (*)(X*)> (memPtr);  // Illegal!

    X x;
    (x.*memPtr) ();
    funPtr (&x);
}

On my machine, this prints:

B::foo() x=B
B::foo() x=A

The B class shouldn't be able to print "x=A"! This happens because member-function pointers carry an extra offset that is added to this before the call, in case multiple inheritance comes into play. Casting loses this offset. So, when calling the casted function pointer, this automatically refers to the first base object, while B is the second, printing the wrong value.

PPS: For even more fun: If we plug in Jarod42's suggestion:

template <typename C, void (C::*M)(), typename Obj>
void AsFunc (Obj* p) {
    (p->*M)();
}

int main () {
    void (*funPtr) (X*) = AsFunc<B, &B::foo, X>;

    X x;
    funPtr (&x);
}

the program correctly prints:

B::foo() x=B

If we look at the disassembly of AsFunc , we see:

c90 <void AsFunc<B, &B::foo, X>(X*)>:
 c90:   48 83 c7 01             add    $0x1,%rdi
 c94:   e9 07 ff ff ff          jmpq   ba0 <B::foo()>

The compiler automatically generated code that adds 1 to the this pointer, such that B::foo is called with this pointing to the B base class of X . To make this happen in the AsFunc function (opposed to buried within main ), I introduced the Obj template parameter which lets the p argument be of the derived type X such that AsFunc has to do the adding.

Answer 4

A) Can I get rid of the compiler warning?

Yes - wrap the member function in a call from a static function

(This is a low-tech variant of @Jarod42's template based answer)

B) To what extent is this code guaranteed to work?

It's not (summarizing @Pete Becker's answer). Until you get rid of the warning.

Here's the jist of what we went with. We kept it simple to minimize disruption to the code. We avoided advanced C++ features to maximize the number of people who can work on the code.

#include <iostream>

class MyClass
{
public:
    MyClass(int a) : memberA(a) {}
    static int myMemberFuncStatic(MyClass *obj)
    {
        return obj->myMemberFunc();
    }   
    int myMemberFunc()
    {
        return memberA;
    }

private:
    int memberA;
};

typedef void(*GENERIC_FUNC_TYPE)(void);
typedef int(*FUNC_TYPE)(MyClass *);

int main(int argc, char*argv[])
{
    int (* staticFunc) (MyClass *) = &MyClass::myMemberFuncStatic;
    MyClass myObject(1);
    std::cout << staticFunc(&myObject) << std::endl;
    // All good so far

    // This is actually legal, for non-member functions (like static functions)
    GENERIC_FUNC_TYPE myStoredFunction = reinterpret_cast<GENERIC_FUNC_TYPE> (staticFunc);  // No compiler warning

    // Reinterpret the fn pointer as the static function
    int (*myExtractedFunction)(MyClass*) = (FUNC_TYPE)myStoredFunction;

    // Call it
    std::cout << myExtractedFunction(&myObject) << std::endl;
}