c++11 expanding parameter pack inside of a lambda

Question

I'm trying to assign a lambda to my std::function that contains 1 call to Test for each type specified in the parameter pack, does anyone know how to do this?

template<typename T>
void Test() {
    
}

template<typename ... Ts>
void Expand() {
    std::function<void(void)> func = [Ts] {
        for (const auto& p : { Ts... })
        Test<Ts...>();
    };
}

int main(int argc, char** argv) {
    Expand<int, char>();
}

I am trying to have this expand to...

Test<int>();
Test<char>();

Answer 1

template<typename ... Ts>

Ts here represents types. Template parameters are types.

[Ts] {
// ...
}

A lambda's capture captures values, and discrete objects rather than types. There's no such thing as a capture of types.

for (const auto& p : { Ts... })

Range iteration iterates over values in some container. A braced initialization list is a list values too, not types.

The major confusion here is the confusion between types and values, in C++.

If the intent is to generate "call to Test for each type specified in the parameter pack", then the simplest solution is to use C++17's fold expression:

template<typename ... Ts>
void Expand() {
    std::function<void(void)> func = [] {
        ( Test<Ts>(), ...);
    };
}

You tagged your question as C++11, but in the year 2022 it is highly likely that your compiler support C++17 as well.

In the unlikely event that you're limited to C++11, a helper throwaway void function can be used as a crutch:

template<typename ...Args>
void ignore_me(Args && ...args)
{
}

template<typename ... Ts>
void Expand() {
    std::function<void(void)> func = [] {
        ignore_me( (Test<Ts>, 0)...);
    };
}