C ++：函数指向具有可变数量参数的函数

Question

I'm trying to figure out a way of how to be able to assign a function pointer to functions with different number of arguments.

I have a while loop which takes a number of different functions as a conditional statement, so instead of writing multiple while loops with exactly the same code inside I'd like to have one with a function pointer. All the functions are of format bool f(...) . I think some code will best illustrate what I mean:

int a, b, c, d;
MyClass* my_class;

typedef bool (MyClass::*my_fun_t)();
my_fun_t my_fun;

if (condition1)
    my_fun = &MyClass::function_one();
else if (condition2)
    my_fun = &MyClass::function_two(a, b);
else if (condition3)
    my_fun = &MyClass::function_three(a, b, c);
else if (condition4)
    my_fun = &MyClass::function_four(a, b, c, d);

while ((my_class->*my_fun)()) 
{ ... }

Now this obviously doesn't work because the functions have different signatures. Is it at all possible to make it work in a similar fashion? Are functoids something I should look at?

Answer 1

You could use std::function<> and std::bind() .

#include <functional>

using std::placeholders::_1;

typedef std::function<bool(MyClass&)> my_fun_t;
my_fun_t my_fun;

if (condition1)
    my_fun = std::bind(&MyClass::function_one, _1);
else if (condition2)
    my_fun = std::bind(&MyClass::function_two, _1, a, b);
else if (condition3)
    my_fun = std::bind(&MyClass::function_three, _1, a, b, c);
else if (condition4)
    my_fun = std::bind(&MyClass::function_four, _1, a, b, c, d);

while (my_fun(my_class)) { ... }

These assumes you will use C++11. If you can't use C++11 but can use TR1, replace all std:: with std::tr1:: . There is also a Boost implementation .

Answer 2

This works for me:

#include <iostream>
#include <cstdarg>

using namespace std;

class MyInterface
{
public:
    virtual bool func(int argc, ...) = 0;
};

class MyImpl : public MyInterface
{
public:
    virtual bool func(int argc, ...);
};

bool MyImpl::func(int argc, ...)
{
    va_list varargs;
    va_start(varargs,argc);
    cout << "Arguments passed:" << endl;
    for(int i = 0; i < argc; ++i)
    {
        // expect double values
        double val = va_arg(varargs,double);
        cout << val << endl;
    }
    va_end(varargs);
    return true;
}

typedef bool (MyInterface::*MyFunc)(int, ...);

int main() {

    MyImpl impl;
    MyInterface* interface = &impl;
    MyFunc pfunc = &MyInterface::func;

    if(!(interface->*pfunc)(2,double(3.14),double(2.72)))
    {
        return 1;
    }
    return 0;
}

Output:

Arguments passed:
3.14
2.72

Obviously you CAN declare and use function pointers for (member-)functions using variable arguments.

Answer 3

You want std::function , a polymorphic function object, and std::bind to create function objects by binding arguments to the parameters of other functors.

If you can't use C++11, then boost::function and boost::bind are equivalent, although slightly more restrictive.

std::function<bool()> my_fun;

if (condition1)
    my_fun = std::bind(&MyClass::function_one, my_class);
else if (condition2)
    my_fun = std::bind(&MyClass::function_two, my_class, a, b);
else if (condition3)
    my_fun = std::bind(&MyClass::function_three, my_class, a, b, c);
else if (condition4)
    my_fun = std::bind(&MyClass::function_four, my_class, a, b, c, d);

while (my_fun()) 
{ ... }

Answer 4

I'm trying to figure out a way of how to be able to assign a function pointer to functions with different number of arguments.

You can't. Function pointers are specific to one function signature. This is entirely logical: how would you invoke such a function? (Yes, C allows invoking functions without specifying in their declaration how many parameters the function has, but this doesn't work in C++ since it subverts the type system.)

Are functoids something I should look at?

Generally yes, but they don't solve this problem.