Question

Given the following class, which simply maps an internal functor f to a function to be run later:

class A {
private:
   int (A::*f)(int);
   int foo(int x) { return x; }
   int bar(int x) { return x*2; }
public:
   explicit A(bool foo=true) { f = foo ? &A::foo : &A::bar; }
   int run(int x) { return (this->*f)(x); }
};

Now say I have another class, B:

class B {
public:
   int foo(int) { return x*x; }
};

And function foo:

int foo(int x) { return 0; }

I know it is not possible to have A assign and run B::foo or foo as their prototypes differ: int (A::*)(int) vs int (B::*)(int) vs int (*)(int).

What I am asking, is their any way to templatize A::f such that it could take any of them?

Answer 1

I am not exactly sure what you are trying to achieve, but you may want to look into:

• Boost.Function – can wrap any kind of callable
• Boost.Bind – can be used to bind a member function to an instance