CODEWITHSUNDEEP
c++boost
Matt Cruikshank
Matt Cruikshank

Reputation: 2968

Passing boost::any to results of boost::bind

I'm trying to figure out how to write this function:

template <typename Bound>
Bound::result_type callFromAnyList(Bound b, list<any> p)
{
}

Then, if I had some function:

double myFunc(string s, int i)
{
    return -3.0;
}

I could call it by doing something like this:

list<any> p;
p.push_back((string)"Hello");
p.push_back(7);
double result = callFromAnyList(bind(myFunc, _1, _2), p);

Is it possible to write something like my callFromAnyList function? Can you inspect the result type and the parameter types from the type returned from bind? And then call any_cast<P1>(*p.begin()), etc? I've tried to understand the bind code, but it's a little hard to follow, and it doesn't appear as though they wrote it with inspection in mind.

Upvotes: 1

Views: 799

Answers (2)

Matt Cruikshank
Matt Cruikshank

Reputation: 2968

I ended up doing this for now -

void invoke(void (f)(), list<any>& params)
{
    f();
}

template <typename R>
void invoke(R (f)(), list<any>& params)
{
    params.push_front(f());
}

template <typename T0>
void invoke(void (f)(T0), list<any>& params)
{
    T0 t0 = any_cast<T0>(*params.begin()); params.pop_front();
    f(t0);
}

template <typename R, typename T0>
void invoke(R (f)(T0), list<any>& params)
{
    T0 t0 = any_cast<T0>(*params.begin()); params.pop_front();
    params.push_front(f(t0));
}

template <typename T0, typename T1>
void invoke(void (f)(T0, T1), list<any>& params)
{
    T0 t0 = any_cast<T0>(*params.begin()); params.pop_front();
    T1 t1 = any_cast<T1>(*params.begin()); params.pop_front();
    f(t0, t1);
}

template <typename R, typename T0, typename T1>
void invoke(R (f)(T0, T1), list<any>& params)
{
    T0 t0 = any_cast<T0>(*params.begin()); params.pop_front();
    T1 t1 = any_cast<T1>(*params.begin()); params.pop_front();
    params.push_front(f(t0, t1));
}

template <typename T0, typename T1, typename T2>
void invoke(void (f)(T0, T1, T2), list<any>& params)
{
    T0 t0 = any_cast<T0>(*params.begin()); params.pop_front();
    T1 t1 = any_cast<T1>(*params.begin()); params.pop_front();
    T2 t2 = any_cast<T2>(*params.begin()); params.pop_front();
    f(t0, t1, t2);
}

template <typename R, typename T0, typename T1, typename T2>
void invoke(R (f)(T0, T1, T2), list<any>& params)
{
    T0 t0 = any_cast<T0>(*params.begin()); params.pop_front();
    T1 t1 = any_cast<T1>(*params.begin()); params.pop_front();
    T2 t2 = any_cast<T2>(*params.begin()); params.pop_front();
    params.push_front(f(t0, t1, t2));
}

I'm lacking the full power of boost::bind - like, I can't handle method pointers - but I also realized that by doing this, I've got a stack processor. I can keep invoking methods that operator on the parameters on the stack.

Upvotes: 0

Johannes Schaub - litb
Johannes Schaub - litb

Reputation: 507273

As you updated your concerns in the comment sections, here the answer. Just getting the return type of a function is possible:

template<typename>
struct return_of;

template<typename R>
struct return_of<R(*)()> {
    typedef R type;
};

template<typename R, typename P1>
struct return_of<R(*)(P1)> {
    typedef R type;
    typedef P1 parameter_1;
};

void foo(int);

template<typename Func>
typename return_of<Func>::parameter_1 bar(Func f) {
    return 42;
}

// call: bar(foo);

I guess you see what this comes down to :) You can use boost function types which already has solved it: http://www.boost.org/doc/libs/1_37_0/libs/function_types/doc/html/index.html

Upvotes: 1

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