How can I access the types of a lambda in c++0x?

Question

How is it possible to access the types of the parameters of a lambda function in c++? The following does not work:

template <class T> struct capture_lambda {
};

template <class R, class T> struct capture_lambda<R(T)> {
    static void exec() {
    }
};

template <class T> void test(T t) {
    capture_lambda<T>::exec();
}

int main() {
    test([](int i)->int{ return 0; });
}

The above does not compile, because the compiler chooses the template prototype instead of the specialization.

Is there a way to do the above?

What I am actually trying to achieve is this: I have a list of functions and I want to select the appropriate function to invoke. Example:

template <class T, class ...F> void exec(T t, F... f...) {
    //select the appropriate function from 'F' to invoke, based on match with T.
}

For example, I want to invoke the function that takes 'int':

exec(1, [](char c){ printf("Error"); }, [](int i){ printf("Ok"); });

I'll ask for a little wider view: What problem are you trying to solve using the `exec` function? Sounds a bit ugly. — GManNickG, Nov 05 '10 at 16:04
Overload resolution on arbitrary function objects types... I can't think of a way to do that. If they would export their param types, it can be done. Otherwise, I've no clue. — Johannes Schaub - litb, Nov 05 '10 at 17:11

score 2 · Accepted Answer · answered Nov 08 '10 at 07:08

This isn't possible, lambda functions are syntactic sugar for creating function objects not actual functions. This means that the template is accepting a class, and classes don't have the concept of argument type.

Also keep in mind that a general function object can have any number of overloaded operator()s.

How can I access the types of a lambda in c++0x?

1 Answers1