折叠表达式的结合性

Question

N4191 向 C++ 提出了折叠表达式。那里的定义是

(args + ...)

是左折（即 (((a0 + a1) + a2) + ...)，

(... + args)

是右折叠（即(... + (a8 + (a9 + a10)))。但是，修改后的论文N4295颠倒了左右一元折叠的定义。

问题：原理是什么？从左到右评估 (args + ...) 似乎更直观（至少当您习惯从左到右的字母表时）。

Answer 1

我不能代表提案，但新的、互换的定义对我来说似乎很自然。我的理由是 (... + args) 是左折叠的子表达式，而 (args + ...) 是右折叠的子表达式。事实上，前者是最后一段，后者是表达式的开始段（我可能使用的术语不正确）。

以下是我将如何从语法中说明折叠的扩展：

左折

                     (... + args)
             (... + args) + a999)
     (... + args) + a998) + a999)

((...((a0 + a1) + a2)...) + a999)

右折

(args + ...)
(a0 + (args + ...)
(a0 + (a1 + (args + ...)

(a0 + (...(a997 + (a998 + a999))...))

Answer 2

根据@cpplearner 的评论，这里有一些来自 std-discussion 的考古学

2015 年 2 月 4 日星期三上午 1:30，@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?

@RichardSmith 的回答

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:

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

(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 (...))).

折叠表达式的结合性

Associativity of fold-expressions

c++

language-lawyer

c++17

associativity

fold-expression