CODEWITHSUNDEEP
c++templatesboost
vintitres
vintitres

Reputation: 63

How to create function template signatures?

I have such template functions:

template<class R> list<R> f(const boost::function<R()>&);
template<class R, class A0> list<R> f(const boost::function<R(T0)>&, list<A0>);
template<class R, class A0, class A1> list<R> f(const boost::function<R(T0)>&, list<A0>, list<A1>);

To run one of them i need to write for example:

int one() { return 1; }
int inc(int x) { return x + 1; }

list<int> l;

f<int>(one);
f<int, int>(inc, l);

And my goal is to just write:

f(one);
f(inc, l);

I heard that this is possible by some kind of template signature specialization, but I can't figure out how.

Upvotes: 4

Views: 214

Answers (1)

Tom Kerr
Tom Kerr

Reputation: 10720

Without C++11, you can't get away from specifying a return type of a function.

template<typename R, typename F>
    R bar(F func) {
        return func();
    }

bar<int>(foo);

With the new C++11 features you can though.

template<typename F>
    auto baz(F func) -> decltype(func()) {
        return func();
    }

baz(foo);

You can template the function/functor as a parameter, instead of trying to specify that it has to be a boost::function.

void zero() {cout << "zero" << endl;}
void one(int a) {cout << "one" << endl;}
void two(int a, int b) {cout << "two" << endl;}

template<typename F>
    void f(const F &func) {
        func();
    }

template<typename F, typename T0>
    void f(const F &func, T0 t0) {
        func(t0);
    }

template<typename F, typename T0, typename T1>
    void f(const F &func, T0 t0, T1 t1) {
        func(t0, t1);
    }

This lets you pass in the function pointer rather simply.

f(zero);
f(one, 1);
f(two, 1, 2);

If you need to actually use functions or bind you can pass that in to the same interface.

// without specifying the function
f(boost::bind(zero));
f(boost::bind(one, _1), 1);
f(boost::bind(two, _1, _2), 1, 2);

// or by specifying the object
boost::function<void()> f0        = boost::bind(zero);
boost::function<void(int)> f1     = boost::bind(one, _1);
boost::function<void(int,int)> f2 = boost::bind(two, _1, _2);
f(f0);
f(f1, 1);
f(f2, 1, 2);

As well as with a functor, which is typical for passing in strict weak ordering behavior to the standard containers.

struct zoobies {
    void operator()() const {}
};

f(zoobies());

It doesn't have to check the type of what you pass in to it, only that it satisfies the interface. This is one of the reasons C++ templates are typically much more powerful than generics in other languages.

And for completeness... If you actually did want to restrict it to boost::function, here is an example.

template<typename T>
    void p(const boost::function<T> &func) {
        func();
    }

template<typename T, typename A0>
    void p(const boost::function<T> &func, A0 a0) {
        func(a0);
    }


boost::function<void()> f0(zero);
p(f0);

boost::function<void(int)> f1(one, _1);
p(f1, 1);

Update:

void foo() {cout << "zero" << endl;}
void foo(int a) {cout << "one" << endl;}
void foo(int a, int b) {cout << "two" << endl;}

boost::bind works out of the box, with this, although the raw function pointers have more of a problem. foo is ambiguous there.

f( (void(*)())        foo      );
f( (void(*)(int))     foo, 1   );
f( (void(*)(int,int)) foo, 1, 2);

If you fully specify the function pointer then it works, though that isn't what anyone wants to do.

With boost::bind as evidence, You should be able to determine the arity from the calling convention of f. If I get some time today I'll play with it.

Upvotes: 2

Related Questions