How to expand into initializer list with fold expressions?

Question

I would like to insert as many zeros into a vector as there are arguments to a variadic templated function (i.e. number of arguments zeros into a vector). I am trying to use fold expressions to achieve this and it works when using (vec.push_back(zeroHelper(args)), ...);.

What I do not understand: Why does it not work when I try to initialize the vector directly by "unfolding" into an initializer list like follows:
std::vector<int> vec = { (zeroHelper(args), ...) };?

Full source code:

template <typename T>
T zeroHelper (T a) { return T{0}; }

template<typename ...Args>
void printer(Args&&... args) {
    std::vector<int> vec; // = { (zeroHelper(args), ...) };
    (vec.push_back(zeroHelper(args)), ...);
    for (auto i : vec) {
        std::cout << i << '\n';
    }
}

int main()
{
    printer(1, 2, 3, 4);
    return 0;
}

And here's the source on OnlineGDB.

Expected output:

0
0
0
0

Output with the initializer list approach:

0

Why?

Answer 1

Because in

std::vector<int> vec = { (zeroHelper(args), ...) };

the parentheses return only one element, the last one; the comma operator discard the precedings.

You should write

std::vector<int> vec = { zeroHelper(args) ... };

to maintains all elements.

In

(vec.push_back(zeroHelper(args)), ...);

the parentheses activate the folding, the comma discards all elements except the last one but the push_back() operation is applied to vec for every args... elements.

Observe also that, using the discarding propriety of the comma operator, you don't need zeroHelper() at all

You can write

std::vector<int> vec { ((void)args, 0)... };

or

(vec.push_back(((void)args, 0)), ...);