Member-Function Pointers With Default Arguments

Question

I am trying to create a pointer to a member function which has default arguments. When I call through this function pointer, I do not want to specify an argument for the defaulted argument. This is disallowed according to the standard, but I have never before found anything that the standard disallowed that I could not do in some other conformant way. So far, I have not found a way to do this.

Here is code illustrating the problem I'm trying to solve:

class MyObj
{
public:
 int foo(const char* val) { return 1; }
 int bar(int val = 42) { return 2; }
};

int main()
{
 MyObj o;

 typedef int(MyObj::*fooptr)(const char*);
 fooptr fp = &MyObj::foo;
 int r1 = (o.*fp)("Hello, foo.");

 typedef int(MyObj::*barptr)(int);
 barptr bp1 = &MyObj::bar;
 int r2 = (o.*bp1)(); // <-- ERROR: too few arguments for call

 typedef int (MyObj::*barptr2)();
 barptr2 bp2 = &MyObj::bar; // <-- ERROR: Can't convert from int(MyObj::*)(int) to int(MyObj::*)(void)
 int r3 = (o.*bp2)();
    return 0;
}

Any ideas on how to do this in conformant C++ if I do not want to specify any values for the defaulted arguments?

EDIT: To clarify the restrictions a bit. I do not want to specify any default arguments either in the call or in any typedef. For example, I do not want to do this:

typedef int(MyObj::*barptr)(int = 5);

...nor do I want to do this:

typedef int(MyObj::*barptr)(int);
...
(o.barptr)(5);

@Goz: Please read the post again. I'm trying to use a function pointer. — John Dibling
– John Dibling, Commented Feb 8, 2010 at 22:30
The reason default arguments work is the compiler can see it. With just function pointers, any information like defaults is gone completely. You have to specify it, or make a wrapper. — GManNickG
– GManNickG, Commented Feb 8, 2010 at 22:33
@GMan: Right, that's exactly the problem I'm having. I thought there might have been a novel solution I hadn't thought of since, as I said, I've never found anything I couldn't do in standard C++. — John Dibling
– John Dibling, Commented Feb 8, 2010 at 22:41
@John: Nope, you simply need to specify the arguments. Like Andrey said, function pointers have zero relation to the function they bind to. — GManNickG
– GManNickG, Commented Feb 8, 2010 at 22:46

AnT stands with Russia · Accepted Answer · 2014-02-14 14:15:05Z

20

It would be rather strange to expect the function pointers to work the way you expect them to work in your example. "Default argument" is a purely compile-time concept, it is a form of syntactic sugar. Despite the fact that default arguments are specified in the function declaration or definition, they really have nothing to do with the function itself. In reality default arguments are substituted at the point of the call, i.e. they are handled in the context of the caller. From the function's point of view there's no difference between an explicit argument supplied by the user or a default one implicitly supplied by the compiler.

Function pointers, on the other hand, are run-time entities. They are initialized at run time. At run-time default arguments simply don't exist. There's no such concept as "run-time default arguments" in C++.

Some compilers will allow you to specify default arguments in function pointer declaration, as in

void foo(int);

int main() {
   void (*pfoo)(int = 42) = foo;
   pfoo(); // same as 'pfoo(42)'
}

but this is not standard C++ and this does not appear to be what you are looking for, since you want the "default argument " value to change at run time depending on the function the pointer is pointing to.

As long as you want to stick with genuine function pointers (as opposed to function objects, aka functors) the immediate workaround would be for you to provide a parameter-less version of your function under a different name, as in

class MyObj 
{ 
public: 
  ...
  int bar(int val = 42) { return 2; } 
  int bar_default() { return bar(); }
}; 

int main() 
{ 
  MyObj o; 

  typedef int (MyObj::*barptr2)(); 
  barptr2 bp2 = &MyObj::bar_default;
  int r3 = (o.*bp2)(); 
  return 0; 
}

This is, of course, far from elegant.

One can actually argue that what I did above with bar_default could have been implicitly done by the compiler, as a language feature. E.g. given the class definition

class MyObj 
{ 
public: 
  ...
  int bar(int val = 42) { return 2; } 
  ...
};

one might expect the compiler to allow the following

int main() 
{ 
  MyObj o; 

  typedef int (MyObj::*barptr2)(); 
  barptr2 bp2 = &MyObj::bar;
  int r3 = (o.*bp2)(); 
  return 0; 
}

where the pointer initialization would actually force the compiler to implicitly generate an "adapter" function for MyObj::bar (same as bar_default in my previous example), and set bp2 to point to that adaptor instead. However, there's no such feature in C++ language at this time. And to introduce something like that would require more effort than it might seem at the first sight.

Also note that in the last two examples the pointer type is int (MyObj::*)(), which is different from int (MyObj::*)(int). This is actually a question to you (since you tried both in your example): how would you want it to work? With an int (MyObj::*)() pointer? Or with a int (MyObj::*)(int) pointer?

edited Feb 14, 2014 at 14:15

answered Feb 8, 2010 at 22:37

AnT stands with Russia

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5 Comments

John Dibling Over a year ago

Re: "Default argument" is a purely compile-time concept -- true, but I was hoping that since the compiler knows the default values in the same translation unit where I call through the pointer, there would be some way to get it to cough up the info.

GManNickG Over a year ago

@John: And how can the compiler know what the function pointer will point to? It's not bound until run-time; and C++ compilers don't have to do static-analysis.

AnT stands with Russia Over a year ago

@John: I'd say that in any case when the use of function pointers is actually justified, this would be prohibitively expensive. The only reasonable way to implement this would be to stuff all this information into the pointer itself, i.e. use a "fat" pointer. This just isn't done that way in C++. If you need a "fat" pointer, use functor instead.

John Dibling Over a year ago

@AndreyT: I suspect my estimation of when the use of function pointers is justified is looser than yours. :)

AnT stands with Russia Over a year ago

@John Dibling: In your case it is "easy" for the compiler to figure out what the arguments are supposed to be only because the pointer bindings are obvious to the compiler. But normally, when the binding are so obvious, there's no need for pointers at all (as in your artificial example). Once the bindings become truly run-time, it simply becomes impossible for the compiler to correctly determine the default argument values (unless it stuffs these values into the pointer itself or uses the approach I described at the end of my answer).

villintehaspam · Accepted Answer · 2010-02-08 22:37:23Z

4

You could create functors instead of function pointers of course.

struct MyFunctor {
    int operator() {
        return myobj.bar();
    }

    MyFunctor(MyObj &obj) : myobj(obj) {}
    MyObj &myobj;
};

then:

MyFunctor myfunc(o);
myFunctor();

answered Feb 8, 2010 at 22:37

villintehaspam

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9 Comments

Manuel Over a year ago

How is using this functor different from calling o.bar() directly?

villintehaspam Over a year ago

@Manuel: The usage wouldn't be so trivial as to create a functor, then immediately call it. Rather you would then give the functor object to some other object that wouldn't know about what MyObj is or what the bar function does. It would only know that it has something with a ()-operator on it.

John Dibling Over a year ago

@Manuel: It's not. Which is why I only said this gave me an idea. This is not the solution I was looking for.

Manuel Over a year ago

@villintehaspam - And can't you think of a way to automatize this with boost::bind or something of the kind?

villintehaspam Over a year ago

@Manuel: Since a boost::bind would use a member function pointer, I'd assume that boost::bind would not work.

|

user269044 · Accepted Answer · 2010-02-08 22:36:06Z

2

This is not possible given the constraints. Your options are:

Using function wrappers.
Using Functors.

Check out Boost for some handy tools to simplify this.

answered Feb 8, 2010 at 22:36

user269044

1485 bronze badges

1 Comment

Manuel Over a year ago

I think that there's nothing in Boost that will prevent him from having to specify a value for the default argument one way or another. Check this related thread: stackoverflow.com/questions/2221832/…

Waqar · Accepted Answer · 2020-07-06 06:04:44Z

Task: Suppose you have the following:

class Thing {
    public:
        void foo (int, double = 3.14) const {std::cout << "Thing::foo(int, double = 3.14) called.\n";}
        void goo (int, double = 1.5) const {std::cout << "Thing::goo(int, double = 1.5) called.\n";}
};

void function1 (const Thing& thing, int a, int b, double c) {
    // Code A
    thing.foo(a,c);
    // Code B
    thing.foo(b);
    // Code C
}

void function2 (const Thing& thing, int a, int b, double c) {
    // Code A
    thing.goo(a,c);
    // Code B
    thing.goo(b);
    // Code C
}

We want to write a helper function to capture function1 and function2 so that the repeated codes A, B, C need not be written twice.

The following will not compile:

class Thing {
    public:
        void foo (int, double = 3.14) const {std::cout << "Thing::foo(int, double = 3.14) called.\n";}
        void goo (int, double = 1.5) const {std::cout << "Thing::goo(int, double = 1.5) called.\n";}
};

void functionHelper (const Thing& thing, int a, int b, double c, void (Thing::*f)(int, double) const) {
    // Code A
    (thing.*f)(a,c);
    // Code B
//  (thing.*f)(b);  // Won't compile.  Too few arguments passed to (thing.*f), which expects (int, double).
    // Code C   
}

void function1 (const Thing& thing, int a, int b, double c) {
    functionHelper (thing, a, b, c, &Thing::foo);
}

void function2 (const Thing& thing, int a, int b, double c) {
    functionHelper (thing, a, b, c, &Thing::goo);
}

First solution (overload of Thing::foo and Thing::goo):

#include <iostream>

class Thing {
    public:
        void foo (int, double = 3.14) const {std::cout << "Thing::foo(int, double = 3.14) called.\n";}
        void foo_default (int a) const {
                std::cout << "Thing::foo_default(int) called.\n";
                foo(a);
        }
        void goo (int, double = 1.5) const {std::cout << "Thing::goo(int, double = 1.5) called.\n";}
        void goo_default (int a) const {
                std::cout << "Thing::goo_default(int) called.\n";
                goo(a);
        }
};

void functionHelper (const Thing& thing, int a, int b, double c,
        void (Thing::*f)(int, double) const, void (Thing::*g)(int) const) {
    // Code A
    (thing.*f)(a,c);
    // Code B
    (thing.*g)(b);  // This will compile now, since (thing.*g) expects int only as argument.
    // Code C   
}

void function1 (const Thing& thing, int a, int b, double c) {
    functionHelper (thing, a, b, c, &Thing::foo, &Thing::foo_default);
}

void function2 (const Thing& thing, int a, int b, double c) {
    functionHelper (thing, a, b, c, &Thing::goo, &Thing::goo_default);
}

int main() {
    Thing thing;
    function1 (thing, 2, 5, 1.8);
    std::cout << '\n';
    function2 (thing, 2, 5, 1.8);
}

Output:

Thing::foo(int, double = 3.14) called.
Thing::foo_default(int) called.
Thing::foo(int, double = 3.14) called.

Thing::goo(int, double = 1.5) called.
Thing::goo_default(int) called.
Thing::goo(int, double = 1.5) called.

Second solution (Wrap Thing::foo and Thing::goo into function objects):

#include <iostream>
#include <memory>

class Thing {
    public:
        void foo (int, double = 3.14) const {std::cout << "Thing::foo(int, double = 3.14) called.\n";}
        void goo (int, double = 1.5) const {std::cout << "Thing::goo(int, double = 1.5) called.\n";}
        
        class FooOrGoo {
            public: 
                void operator()(const Thing& thing, int a) const {helper1 (thing, a);}
                void operator()(const Thing& thing, int a, double b) {helper2 (thing, a, b);}
                virtual ~FooOrGoo() {std::cout << "Thing::FooOrGoo object destroyed.\n";}
            private:
                virtual void helper1 (const Thing& thing, int a) const = 0;
                virtual void helper2 (const Thing& thing, int a, double b) const = 0;
        };

        class Foo : public FooOrGoo {
            virtual void helper1 (const Thing& thing, int a) const override {thing.foo(a);}
            virtual void helper2 (const Thing& thing, int a, double b) const override {thing.foo(a, b);}
        };

        class Goo : public FooOrGoo {
            virtual void helper1 (const Thing& thing, int a) const override {thing.goo(a);}
            virtual void helper2 (const Thing& thing, int a, double b) const override {thing.goo(a, b);}
        };
};

void functionHelper (const Thing& thing, int a, int b, double c, std::unique_ptr<Thing::FooOrGoo> f) {
    // Code A
    (*f)(thing, a,c);
    // Code B
    (*f)(thing, b);
    // Code C   
}

void function1 (const Thing& thing, int a, int b, double c) {
    functionHelper (thing, a, b, c, std::unique_ptr<Thing::Foo>(new Thing::Foo));  // 'std::make_unique<Thing::Foo>());' is not supported by GCC 4.8.1.
}

void function2 (const Thing& thing, int a, int b, double c) {
    functionHelper (thing, a, b, c, std::unique_ptr<Thing::Goo>(new Thing::Goo));  // 'std::make_unique<Thing::Goo>());' is not supported by GCC 4.8.1.
}

int main() {
    Thing thing;
    function1 (thing, 2, 5, 1.8);
    std::cout << '\n';
    function2 (thing, 2, 5, 1.8);
}

Output:

Thing::foo(int, double = 3.14) called.
Thing::foo(int, double = 3.14) called.
Thing::FooOrGoo object destroyed.

Thing::goo(int, double = 1.5) called.
Thing::goo(int, double = 1.5) called.
Thing::FooOrGoo object destroyed.

Which solution do you think is better? I think the second one is more elegant, but there are more lines of code (I couldn't do it without polymorphism).

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