How to declare two functions taking each other's signature as argument?

Question

Is it possible to emulate something like this:

typedef boost::function<void(A)> B;
typedef boost::function<void(B)> A;

The main goal is to be able to write code like this (in pseudo-c++):

void a_(B b) {
  // ...
  b(a_);
}
void b_(A a) {
  // ...
  f(boost::bind(a, b_));
}

f(boost::bind(a_, b_));

Answer 1

Not directly possible with typedefs; wherever a typedef is used, it is equivalent to the original type, so if you write

typedef boost::function<void(A)> B;
typedef boost::function<void(B)> A;

then B would be equivalent to boost::function<void(A)>, which is equivalent to boost::function<void(boost::function<void(B)>)>, and so on, until you get

boost::function<void(boost::function<void(boost::function<void(...)>)>)>

, which is a type of infinite length.

You can, however, define (at least) one of the two types as a struct or class:

struct A;
typedef boost::function<void(A)> B;
struct A
{
    B b;
    A(B b) : b(b) {}

    // optional:
    void operator() (A a) { b(a); }
};

You might need to add more constructors and/or a conversion operator to make the type behave completely "transparently", or you could just access the struct explicitly.