How does folding over comma work?

Question

How does the following line unfold?

template <class... Ts>
void print_all(std::ostream& os, Ts const&... args) {
    (void(os << args), ...);
}

Applying the rule,

Unary right fold (E op ...) becomes E₁ op (... op (E_N-1 op E_N))

provided by cppreference,

E  = void(os << args)
op = , 

Then the expansion becomes

void(os << args[0], ..., (args[N-3], (args[N-2], args[N-1])) )

?

How about

v.push_back(args), ...

Does it become

v.push_back(args[0], (args[1], ..., (args[N-2], args[N-1])))

Both the expansion and the parenthesis are confusing. Would someone explain?

Answer 1

You have to unpack the entire expression that contains the parameter pack. Not just the parameter pack. Just following the rules:

Unary right fold (E op ...) becomes E₁ op (... op (E_N-1 op E_N))

You're right that op is , and E is void(os << args), where args is the pack, but E_i isn't just args_i, it's the whole expression void(os << args#i). So:

(void(os << args), ...);

becomes (using [] for convenience):

void(os << args[0]), void(os << args[1]), void(os << args[2]), ..., void(os << args[N-1]);

which is equivalent to:

os << args[0];
os << args[1];
...
os << args[N-1];

Similarly, (v.push_back(args), ...) (the parentheses are required) would expand as:

v.push_back(args[0]), v.push_back(args[1]), ..., v.push_back(args[N-1]);

---

Note that you could also write this example as a binary left fold:

(os << ... << args);

which would expand as:

((((os << args[0]) << args[1]) ... ) << args[N-1]);