Associativity of fold-expressions

Question

N4191 proposed fold-expressions to C++. The definition there was that

(args + ...)

is a left-fold (i.e. (((a0 + a1) + a2) + ...),

Answer 1

From the comment by @cpplearner, here's some archeology from std-discussion

On Wed, Feb 4, 2015 at 1:30 AM, @T.C. wrote:

In N4295, which was actually voted into the standard,

(... op e) is a unary left fold;

(e op ...) is a unary right fold;

In N4191, however,

(e op ...) is called a left fold.

(... op e) is called a right fold.

Why the 180-degree turn?

And the answer by @RichardSmith

The form in the original paper was simply a typo. Here are some reasons why the definition that was voted into the standard is the correct one:

1. In the standard's formulation, (e op ...) has subexpressions of the form (e_i op ). It does not have subexpressions of the form ( op e_i). This is consistent with all other pack expansions, where the expansion comprises repeated instances of the pattern.

2. (e op ... op eN), where eN is a non-pack, must have eN as the innermost operand in order to be useful -- that is, it must be (e1 op (e2 op (e3 op (... op eN)...))), not (...(((e1 op e2) op e3) op ...) op eN) -- and vice versa for (e0 op ... op e). This allows, for instance, (string() + ... + things) and (std::cout << ... << things) to work. For consistency, (e op ...) must also be (e1 op (e2 op (...))).