减少元运算符不一致



当我们检查reduce函数时:

my $result = reduce &reduction_procedure, @array;

我们以以下简单的内部工作规则结束:

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)

这个简单的规则也适用于归约元运算符 []。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 我们观察到不一致:

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)

不一致的是,从 Step.2 及以后,我们不能将上一步的结果用作后续 reduce 操作的第一个参数, 但相反,我们使用实际值计算一个逻辑中间结果,最后我们在最后一步添加使用"逻辑 AND" 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)

有点说我们有一个"元还原"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) ?

这不是不一致,而是运算符关联性在 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<运算符具有链关联性,因此它有效地将其视为:>

(1 < 2) and (2 < 3) and (3 < 4)

这是真的。

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