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]);

• To be more precise, should the unpacking of the unary case be identical to the binary case with all the parenthesis? (assuming the same unpacking order is applied). – Candy Chiu Aug 10 '17 at 15:12
• 1
  @Barry did you omit some parenthesis in the unary expansion? void(os << args[0]), (void(os << args[1]), (void(os << args[2]), ... (void(os << args[N-2], void(os << args[N-1]))...)); the parenthesis may be important in other scenarios. – Candy Chiu Aug 11 '17 at 14:56
• 1
  @CandyChiu Yes, but they're not important in this scenario, so they don't really help in explaining what happens. They're just... more parentheses. – Barry Aug 11 '17 at 15:08
• 1
  why is casting to void needed? – In78 Jan 21 at 19:07
• 1
  @Silicomancer All folding requires parentheses, it's part of the grammar. – Barry Mar 21 at 18:37

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