c++ - Why doesn't a left fold expression invert the output of a right fold expression? -

i'm taking @ c++17 fold expressions , i'm wondering why following program outputs

4 5 6  4 5 6  

for both of for_each calls

template<typename f, typename... t> void for_each1(f fun, t&&... args) {     (fun (std::forward<t>(args)), ...); }  template<typename f, typename... t> void for_each2(f fun, t&&... args) {     (..., fun (std::forward<t>(args))); }  int main() {      for_each1([](auto i) { std::cout << << std::endl; }, 4, 5, 6);      std::cout << "-" << std::endl;      for_each2([](auto i) { std::cout << << std::endl; }, 4, 5, 6); } 

live example

i thought second fold expression meant output numbers in reverse order

6 5 4 

how come results same?

according § 14.5.3/9

the instantiation of fold-expression produces:

(9.1) — ((e1 op e2) op · · · ) op en unary left fold,

(9.2) — e1 op (· · · op (en-1 op en )) unary right fold,

(9.3) — (((e op e1) op e2) op · · · ) op en binary left fold, and

(9.4) — e1 op (· · · op (en-1 op (en op e))) binary right fold

in each case, op fold-operator, n number of elements in pack expansion parameters, , each ei generated instantiating pattern , replacing each pack expansion parameter ith element.

in code above they're both unary fold expressions , expansion is

template<typename f, typename... t> void for_each1(f fun, t&&... args) {      // unary right fold (fun(args_0) , (fun(args_1) , (fun(args_2) , ...)))     (fun (std::forward<t>(args)), ...); }  template<typename f, typename... t> void for_each2(f fun, t&&... args) {      // unary left fold ((fun(args_0) , fun(args_1)) , fun(args_2)) , ...     (..., fun (std::forward<t>(args)));  } 

so expressions have same evaluation order defined comma operator , output same.

_{credits: friend marco raised original question in first place , gave me chance solve potentially-misleading issue.}