当我们检查reduce函数时:

    my $result = reduce &reduction_procedure, @array;

我们以以下简单规则总结内部工作原理:
    Reduction Rules
    ---------------
    1. For the first pair of objects from our input (@array)
       we apply the reduction procedure (&reduction_procedure) and we get our first result.

    2. We apply the reduction procedure (&reduction_procedure) to the result (object)
       of the previous rule and the next input (object) from our input (@array),
       and we get an intermediate result (object).

    3. We run rule.2 for every of the remaining objects from our input (@array)

这个简单的规则对于还原元运算符[]也同样起作用。例如:
    example.1
    ---------
    say [+] [1,2,3,4]; # result : 10

例如.1,减少规则按原样应用:
    Step.1 use Rule.1 : 1 + 2 = 3     1(first value)     + 2(second value)  = 3
    Step.2 use Rule.2 : 3 + 3 = 6     3(previous result) + 3(current value) = 6
    Step.3 use Rule.2 : 6 + 4 = 10    6(previous result) + 4(current value) = 10

但不适用于以下示例:
    example.2
    ----------
    say [<] 1,2,3,4;   # result : True

例如,2我们观察到一个不一致的地方:
    Step.1 use Rule.1 : 1 < 2 = True    1(first value)         <  2(second value)      = True
    Step.2 use Rule.2 : 2 < 3 = True    True(previous result) &&  True(current result) = True
    Step.3 use Rule.2 : 3 < 4 = True    True(previous result) &&  True(current result) = True(result)

从第2步开始,我们不能将上一步的结果用作后续的reduce操作的第一个参数,
但是相反,我们使用实际值来计算逻辑中间结果,最后,在最后一步中,将“逻辑与”的使用添加为
每个步骤的中间逻辑结果:
    Reduction Result = True && True && True = True (use of logical AND as "meta-reduction" operator)

可以说,我们有一个“减少元数据”的元运算符!

问题:
    1. Is that an inconsistency with a purpose?

    2. Can we exploit this behavior? For instance instead of use of && (logical AND)
       as "meta-reduction" operator can we use || (logical OR) or ?^ (logical XOR) ?

最佳答案

这不是矛盾,而是operator associativity在Raku中如何工作的示例。

写作 :

[<] 1,2,3,4

与写作相同:
1 < 2 < 3 < 4

在大多数语言中,这是行不通的,但是
(1 < 2) and (2 < 3) and (3 < 4)

这是真的。

10-08 07:35