Problem: passing function as argument !!!

Hi, I'm having a problem in pass a function as argument to another function.
This call is made by reference, and the function called is declared into a class.
A example of my problem is:

Code:

---

class ExClass {
      public:
            int func(const int &); //***
};

int ExClass::func(const int &z) { return(z); }

template <class T>
void example(T &func) {}

int main(){
    ExClass f;
    example( f.func );
    return(0);
}

---

When I compile that, I got such error:

In function ‘int main()’:
error: no matching function for call to ‘example(<unresolved overloaded function type>)’
note: candidates are: void example(T&) [with T = int (ExClass::*)(const int&)]


*** It's good to mention that if I declared "func" as "static int func(const int &);" into the "ExClass", I get no error. But I need let "func" be non-static.

What am I supposed to do?
Thank you.

You can't get from a "plain" function pointer to a pointer to member function, or back the other way. (Since a member function has to be called from an object of the class, and has to know about that object, etc., which a "plain" function pointer can't.)

What are you trying to do, really?

>> Hi, I'm having a problem in pass a function as argument to another function.

It can't work because you're passing in a function that takes an implicit pointer to *this, which you haven't supplied, and without an object the function cannot be called, obviously.

I'm intending to use a 'ready' function, Broydn, from 'Numerical Recipes Book". And that function, Broydn, calls aother function by reference. For example:

Code:

---

template <class T>
void Broydn( T &func ) {}

---

When my member function, "func", was declared as 'static', so I could use the syntax:

Code:

---

int main(){
    ExClass f;
    Broydn( f.func );
    return(0);
}

---

But, I can't understand why, if I declare my member function as non-static it doesn't work.

Thank you.

Quote:

---

Originally Posted by tharnier

...if I declared "func" as "static int func(const int &);" into the "ExClass", I get no error. But I need let "func" be non-static.

---

I don't think you can do this without a static member. I tried a few different things on my own, and it seems you can't take the address of a non-static member function.

Well, I had thought we had covered why: You would need an ExClass object to call your function with -- and Broydn quite simply won't have such an object, and therefore quite simply won't be able to call the function. What is this object ExClass doing for you? (I mean, I had to look up Broyden's method just now, and it appears to be a root-finder.) If you need this class, I suppose you can adjust your root-finder to take an object and a member function pointer, so that your root finder will have the object needed to call the function with.

Quote:

---

Originally Posted by Elkvis

I don't think you can do this without a static member. I tried a few different things on my own, and it seems you can't take the address of a non-static member function.

---

hmmmm, sorry if I was not clear. I think this explain why I couldn't do:

Code:

---

int main(){
    ExClass f, *p=&f;
    p -> func;
...
}

---

because I cant take the address of a non-static member function, am I right?

Quote:

---

Originally Posted by tabstop

Well, I had thought we had covered why: You would need an ExClass object to call your function with -- and Broydn quite simply won't have such an object, and therefore quite simply won't be able to call the function. What is this object ExClass doing for you? (I mean, I had to look up Broyden's method just now, and it appears to be a root-finder.) If you need this class, I suppose you can adjust your root-finder to take an object and a member function pointer, so that your root finder will have the object needed to call the function with.

---

ow yeah, I think you really helped me right here!!!
Broydn is a root finder, you're right. And you let my problem clear; I need that Broydn be able to get an object from my class ExClass.
But how should I do that?

I thought that if I had declared Broydn as "friend" of "ExClass", so Broydn could use a member function even without a object, couldn't??

Two examples.
One using boost, the other without boost.

Code:

---

class ExClass
{
public:
    int func(const int& z); //***
};

int ExClass::func(const int& z) { return z; }

template <typename T>
void example(T& func) { func(); }

template <typename Class_t, typename Fnc_t>
void example2(Class_t& Class, Fnc_t FncPtr) { (Class.*FncPtr)(); }

int main()
{
    ExClass f;
    example( boost::bind(&ExClass::func, f) );
    example2(f, &ExClass::func);
    return 0;
}

---

And don't forget to read this:
https://apps.sourceforge.net/mediawi...arameter_names

Quote:

---

Originally Posted by tharnier

I thought that if I had declared Broydn as "friend" of "ExClass", so Broydn could use a member function even without a object, couldn't??

---

Of course not. That's just silly. Friends has nothing to do with it.
The problem is that the functions are members OF objects. They are not global!
So if you want to call a member function, you need to specify which class instance's member function you want to call!
Thus you need the instance AND the member function pointer.

What you're looking for is a 'functor'. In its crudest form it would work something like this:

Code:

---

class ExClass {
        public:
       
        int func(const int & value)
        {
                cout << "Value: " << value << endl;
                return value;
        }

        struct wrapper
        {
                wrapper(ExClass* this_)
                : this_(this_)
                {        }
       
                int operator ( )(const int & value)
                {
                        return this_->func(value);
                }
               
                ExClass* this_;
        };
       
        wrapper wrap(void)
        {
                return wrapper(this);
        }       
};

template <class T>
void example(T &func)
{
        int value = 10;
        func(value);
}

int main(void) {
        ExClass
                e;
        ExClass::wrapper
                f = e.wrap( );
        example(f);               
}

---

In that case, I'll throw in another, better example:

Code:

---

class ExClass
{
public:
    int func(const int& z); //***
};

int ExClass::func(const int& z) { return z; }

template<typename Class_t, typename Fnc_t, typename Return_t>
class Functor
{
public:
    Functor(Class_t& Class, Fnc_t FncPtr): m_Class(Class), m_FncPtr(FncPtr) { }
    Return_t operator () { return m_FncPtr(); }
    const Return_t operator () const { return m_FncPtr(); }

private:
    Class_t m_Class;
    Fnc_t m_FncPtr;
};

template<typename Return_t, typename Class_t, typename Fnc_t>
Functor<Class_t, Fnc_t, Return_t> FunctorWrapper(Class_t& Class, Fnc_t FncPtr)
{
    return Functor<Class_t, Fnc_t, Return_t>(Class, FncPtr);
}

template<typename T> void example(T& Functor) { Functor(); }

int main()
{
    ExClass f;
    example( FunctorWrapper<int>(f, &ExClass::func) );
    return 0;
}

---

I'm not sure how it handles void return type, however. It may be that it needs extra work. Perhaps a specialization.

>> I'm not sure how it handles void return type, however.

AFAIK, it should handle that automagically. I could be wrong, though.

>> const Return_t

You probably shouldn't return const like that. The user might not want it to be, and anyway it wouldn't work if the type was already const. At any rate, yours is a much better way to handle a functor...I was just being lazy. :p

Quote:

---

Originally Posted by tharnier

I thought that if I had declared Broydn as "friend" of "ExClass", so Broydn could use a member function even without a object, couldn't??

---

"friend" just means that it can access private variables of an object which it still needs to have -- you can't still talk about "the" variable x, it's bob's variable x, or fred's variable x, or joe's variable x, et cetera. Members don't exist by themselves, only as part of other objects (hence the name).

Quote:

---

Originally Posted by Elysia

In that case, I'll throw in another, better example:

---

I think that will work. But I'm just taking my time to understand the code....
Thank you everyone.

Quote:

---

Originally Posted by Sebastiani

>> I'm not sure how it handles void return type, however.

AFAIK, it should handle that automagically. I could be wrong, though.

---

I'm just concerned about returning void from the FncPtr call.

Quote:

---

>> const Return_t

You probably shouldn't return const like that. The user might not want it to be, and anyway it wouldn't work if the type was already const. At any rate, yours is a much better way to handle a functor...I was just being lazy. :p

---

That's why there are two overloads! One for const, one for non-const.
It's critical if we pass the object through a const reference.
So don't use "const" in the type and it'll work ;)

>> That's why there are two overloads! One for const, one for non-const.

What, you expect me to actually read the code? :D

>> I'm just concerned about returning void from the FncPtr call.

Code:

---

template < typename R >
R foo( R ( *f )( ) )
{
        return f( );
}

int bar( void )
{
        return 3114;
}

void qux( void )
{        }

int main( void )
{
        cout << foo( bar ) << endl;
        foo( qux );
        return 0;
}

---

Quote:

---

Originally Posted by Sebastiani

>> That's why there are two overloads! One for const, one for non-const.
What, you expect me to actually read the code? :D

---

Aww, come on. It's not that big and complicated.
So this actually compiles, where you try to return "void"?

>> Aww, come on. It's not that big and complicated.

Oh I know, I should have paid closer attention.

>> So this actually compiles, where you try to return "void"?

Yep. As long as you don't try to explicitly declare a variable of that type, though (which is really too bad because that would have been a useful feature). So in other words, this would not compile if used with void:

Code:

---

template < typename R >
R foo( R ( *f )( ) )
{
        R r = f( );               
        return r;
}

---

Alright, cool, no worries about that then.

Quote:

---

Originally Posted by Elysia

Two examples.
One using boost, the other without boost.

Code:

---

class ExClass
{
public:
    int func(const int& z); //***
};

int ExClass::func(const int& z) { return z; }

template <typename T>
void example(T& func) { func(); }

template <typename Class_t, typename Fnc_t>
void example2(Class_t& Class, Fnc_t FncPtr) { (Class.*FncPtr)(); }

int main()
{
    ExClass f;
    example( boost::bind(&ExClass::func, f) );
    example2(f, &ExClass::func);
    return 0;
}

---

And don't forget to read this:
https://apps.sourceforge.net/mediawi...arameter_names

---

I cant read this because its looked...

And, I cant see but when I am compiling that, its returning a error:

Code:

---

In function ‘void example2(Class_t&, Fnc_t) [with Class_t = ExClass, Fnc_t = int (ExClass::*)(const int&)]’:
16:  instantiated from here
10: error: too few arguments to function

---

What is wrong?

Ah, that's because the function takes an int, but we don't pass one. You can fix it like:
(Class.*FncPtr)(10);
func(10);

Quote:

---

Originally Posted by Elysia

Ah, that's because the function takes an int, but we don't pass one. You can fix it like:
(Class.*FncPtr)(10);
func(10);

---

hmmmm that is what I was trying to avoid...
But I'm trying to use boost::function...

Do you know some really good book to indicate to me??? with this kind of doubt

Oh? Then let's borrow the power of boost::bind a little more:
example( boost::bind(&ExClass::func, f, 10) );
Notice that I put a number in there as well.
Now
func();
Simply will work.
Praised be boost!

What probably will solve my problem is:

Code:

---

ExClass f;
f.func = boost::bind( &ExClass::func, &f );
example( f.func );

---

isn't?

But, do I really need to install all boost library to use boost::bind()???

It doesn't work that way! Just pass what boost::bind returns directly to your function.
You need to download the boost package, yes, but you don't need to install any dlls to use it. However, you can't just break it into pieces since it has dependencies on other headers in the boost distribution.

Code:

---

ExClass f;
example( boost::bind(&ExClass::func, f, 1000) );

---

Quote:

---

Originally Posted by Elysia

It doesn't work that way! Just pass what boost::bind returns directly to your function.
You need to download the boost package, yes, but you don't need to install any dlls to use it. However, you can't just break it into pieces since it has dependencies on other headers in the boost distribution.

Code:

---

ExClass f;
example( boost::bind(&ExClass::func, f, 1000) );

---

---

Ok, and how about the book???
Can you suggest one??
To look up that kind of doubt?

Not much about boost, sadly.
There is one book about boost, I think. Can't remember its name. It's somewhat old and doesn't cover that much, but it's about the only one available or so. A search on the web should reveal more information about that.
Otherwise the boost homepage is your best source of information.

Quote:

---

Originally Posted by Elysia

Not much about boost, sadly.
There is one book about boost, I think. Can't remember its name. It's somewhat old and doesn't cover that much, but it's about the only one available or so. A search on the web should reveal more information about that.
Otherwise the boost homepage is your best source of information.

---

No no, not about boost... about c++, but not too beginer, I say intermediate!!!!

Oh, okay. Then check the books thread.
If you haven't checked out Accelerated C++, then you should probably do that too.
It doesn't cover just the basics; it covers a lot of more and a lot of it is handy stuff!

Quote:

---

Originally Posted by Elysia

Oh, okay. Then check the books thread.
If you haven't checked out Accelerated C++, then you should probably do that too.
It doesn't cover just the basics; it covers a lot of more and a lot of it is handy stuff!

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Ok, thank you all your attention and patient to my problem!!!!