C++ Calling Static Function pointer with Predefined parameters

Question

I have a rather unusual problem... we are tasked with mapping a series of hexadecimal values to function pointers. However, I want to be able to call the function with already predefined parameters. Here is a segment of code that I am trying to get to work:

map<short, void(*)(MyClass, MyClass)> mapping = create_map();


map<short, void(*)(MyClass, MyClass)> create_map()
{
    map<short, void(*)(MyClass, MyClass)>> m;
    m[1] = &function(new MyClass(), new MyClass());
    return m;
}

However, it will only let me put function without parameters into the mapping. How do I store this setup and call it later?

@rerun: I think the problem is fairly well-explained. I'm not sure why the environment is relevant? — Oliver Charlesworth
– Oliver Charlesworth, Commented Feb 25, 2012 at 15:55
Why not just create a class that contains the parameters and the function pointer and store that in the map? — Justin Peel
– Justin Peel, Commented Feb 25, 2012 at 15:57
I wanted to know if lambdas or bind were available on his platform. — rerun
– rerun, Commented Feb 25, 2012 at 16:05
Beware that the design as is leaves the responsibility of releasing the memory to the static function, or you will leak the pointers created in the mapping — David Rodríguez - dribeas
– David Rodríguez - dribeas, Commented Feb 25, 2012 at 16:13

Seth Carnegie · Accepted Answer · 2012-02-25 16:14:14Z

3

If you are using C++11, You can use std::bind:

void function(MyClass* a, MyClass* b) { ... }

typedef map<short, std::function<void()>> funcmap;

funcmap create_map()
{
    funcmap m;
    m[1] = std::bind(function, new MyClass, new MyClass);
    return m;
}

// use:

funcmap mapping = create_map();
mapping[1]();

If not, you can create a struct that stores the arguments and the function pointer. Then write an operator() which will call the function through the function pointer with the arguments it stored:

struct func {
    typedef void(*fptr)(MyClass*, MyClass*);

    func(fptr f, MyClass* m1, MyClass* m2) : ptr(f), arg1(m1), arg2(m2) { }

    void operator()() {
        ptr(arg1, arg2);
    }

    T ptr;
    MyClass* arg1, arg2;
};

void function(MyClass* a, MyClass* b) { ... }

typedef map<short, func> funcmap;

funcmap create_map()
{
    funcmap m;
    m[1] = func(function, new MyClass, new MyClass);
    return m;
}

// use:

map<short, func> mapping = create_map();
mapping[1]();

edited Feb 25, 2012 at 16:14

answered Feb 25, 2012 at 15:59

Seth Carnegie

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4 Comments

Mooing Duck Over a year ago

I feel like that shouldn't compile, as the map is expecting a function taking two parameters, but you give is a functor taking zero. I've heard boost::bind is magic, but this use is quite a stretch...

Seth Carnegie Over a year ago

@MooingDuck fixed, I always forget to do that (did it the other day in the chat too)

Mooing Duck Over a year ago

but this solution cannot store regular function pointers to the type in question. I fear his quest is impossible.

Seth Carnegie Over a year ago

@MooingDuck yes, it is impossible to store a plain function pointer with bound arguments, but I think he just wanted to simulate it (that is, I don't think "a real live function pointer" is part of the requirement)

Oliver Charlesworth · Accepted Answer · 2012-02-25 15:56:53Z

1

If you are using C++11, you can use lambda functions.

If you are using an older version of C++, then you can use functors.

answered Feb 25, 2012 at 15:56

Oliver Charlesworth

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