SICP累积功能
计算机程序的结构和解释 (SICP) 第 2.2.3 节 定义了几个函数：
(accumulate cons nil
(filter pred
(map op sequence)))
使用此函数对斐波那契数列表进行操作的两个示例，even-fibs和list-fib-squares。
Accumulate、Filter 和 Map 函数也在 2.2 节中定义。让我困惑的是为什么作者在这里包含了accumulate。 accumulate 需要 3 个参数：
要应用的二元函数
初始值，用作函数最右边的参数
将应用函数的列表
中的定义将累加应用于列表的示例book：
(accumulate cons nil (list 1 2 3))
=> (cons 1 (cons 2 (cons 3 nil)))
=> (1 2 3)
由于第三个参数是一个列表，因此 (accumulate cons nil some-list) 将仅返回 some-list，在本例中为 (过滤器pred（映射操作序列））是一个列表。
除了与本节中其他类似结构的函数保持一致之外，使用 accumulate 是否还有其他原因？
In Structure and Interpretation of Computer Programs (SICP) Section 2.2.3 several functions are defined using:
(accumulate cons nil
(filter pred
(map op sequence)))
Two examples that make use of this operate on a list of the fibonacci numbers, even-fibs and list-fib-squares.
The accumulate, filter and map functions are defined in section 2.2 as well. The part that's confusing me is why the authors included the accumulate here. accumulate takes 3 parameters:
A binary function to be applied
An initial value, used as the rightmost parameter to the function
A list to which the function will be applied
An example of applying accumulate to a list using the definition in the book:
(accumulate cons nil (list 1 2 3))
=> (cons 1 (cons 2 (cons 3 nil)))
=> (1 2 3)
Since the third parameter is a list, (accumulate cons nil some-list) will just return some-list, and in this case the result of (filter pred (map op sequence)) is a list.
Is there a reason for this use of accumulate other than consistency with other similarly structured functions in the section?
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我确信
accumulate的这两种用法只是说明了这样一个事实：“使用元素来构建列表”可以被视为一个累积过程，就像“将数字相乘以获得一个列表”一样。产品”或“将数字相加以获得总数”都可以。你是对的，积累实际上是一个空操作。（旁白：请注意，如果
filter的输出和accumulate的输入不是列表，这显然是一个更有用的操作；例如，如果它表示延迟生成的顺序。）I'm certain that those two uses of
accumulateare merely illustrative of the fact that "consing elements to construct a list" can be treated as an accumulative process in the same way that "multiplying numbers to obtain a product" or "summing numbers to obtain a total" can. You're correct that the accumulation is effectively a no-op.(Aside: Note that this could obviously be a more useful operation if the output of
filterand input ofaccumulatewas not a list; for example, if it represented a lazily generated sequence.)