是否可以展开可变参数（lambda）模板并将这些函数的返回值传递给另一个可变参数函数？

Question

So I wonder given a variadic template function like following:

template<typename...Fs>
parse(int x, Fs...funcs);

Where we ensure (through C++20 concept) that is convertible to std::function<double(int)> . Could we use unfold it into another functions argument, like passing to following one:

template<typename...Ts>
test(Ts...args);

Where again we ensure all Ts is and must be able to convert double .

What is expecting is, suppose a , b , c are lambda expressions, calling parse(12, a, b, c) is equivalent to calling test(a(12), b(12), c(12)) .

---

Something I tried is like following

template<typename...Fs>
parse(int x, Fs...func) {
    return test(..., funcs(x));
}

But didn't work. It seems like that most unfolding expression examples puts a function within the expression, so does this means unfold expression by itself can't return as a list of arguments. Instead, perhaps it could only return as single value?

If so, is there any workaround like first construct it into an array like (..., void(arr[i++]=args)) , but I couldn't find ways to expand it back to argument pack. (I would want to expand arguments back to argument pack as I want to reuse other functions within my library instead of writing another one that takes either array or initializer list as argument.)

Answer 1

Syntax would be:

template <typename... Fs>
auto parse(int x, Fs... func)
{
    return test(funcs(x)...);
}