为什么对 `defmacro` 的调用计算结果为 None？

Question

我正在编写以下代码：

(require [hy.contrib.walk [let]])


(defn maybe? [command-text]
  (let [splitted (.split command-text " ")]
    (= (get splitted 0) "maybe")))


(defn if? [command-text]
  (let [splitted (.split command-text " ")]
    (+ (get splitted 0) "if")))

...直到我意识到我在做一些重复的事情，所以我想分解出模式：

(import [hy [HySymbol]])


(defmacro command-dispatcher [command-type]
  `(defn ~(HySymbol (+ command-type "?")) [command-text]
     (let [splitted (.split command-text " ")]
       (= (get splitted 0) ~command-type))))

但是，如果我在 HyREPL 中计算 (command-dispatcher "maybe")，我会得到 None。

=> (command-dispatcher "maybe")
def is_maybe(command_text):
    _hyx_letXUffffX3 = {}
    _hyx_letXUffffX3['splitted'] = command_text.split(' ')
    return _hyx_letXUffffX3['splitted'][0] == 'maybe'


None

这很奇怪，因为宏应该 return 一个 HyExpression，而不是 None。我错过了什么？

Answer 1

你的宏不会return任何东西，但会定义一个函数，正如你在这里看到的那样

(assert (not (in "is_maybe" (dir))))
(command-dispatcher "maybe")
(assert (in "is_maybe" (dir)))

您的代码中的一个问题是您正在使用 let，根据 documentation, here is a possible way to rewrite it using setv，它不再可用：

(defmacro command-dispatcher [command-type]
  `(defn ~(HySymbol (+ command-type "?")) [command-text]
       (setv splitted (.split command-text " "))
       (= (get splitted 0) ~command-type)))

然后您可以使用is_maybe（或maybe?，这是语法糖）调用此函数，例如

(command-dispatcher "maybe")

(print (maybe? "foo"))
(print (maybe? "maybe foo"))

将打印

False
True

为什么对 `defmacro` 的调用计算结果为 None？

Why are calls to `defmacro` evaluating to None?

macros

hy