匿名函数简写 [英] Anonymous function shorthand

问题描述

有些东西我不明白匿名函数使用短符号＃（..）

There's something I don't understand about anonymous functions using the short notation #(..)

以下工作：

REPL>  ((fn [s] s) "Eh")
"Eh"

不会：

REPL>  (#(%) "Eh")

这样工作：

REPL> (#(str %) "Eh")
"Eh"

< t了解是为什么（＃（％）Eh）不起作用，同时我不需要使用 fn [s] s）Eh）

What I don't understand is why (#(%) "Eh") doesn't work and at the same time I don't need to use str in ((fn [s] s) "Eh")

它们都是匿名函数，它们都有一个参数。为什么简写符号需要一个函数而其他符号不需要？

They're both anonymous functions and they both take, here, one parameter. Why does the shorthand notation need a function while the other notation doesn't?

推荐答案

#(...)

是

(fn [arg1 arg2 ...] (...))

b $ b

（其中argN的数量取决于正文中有多少％N）。所以当你写：

(where the number of argN depends on how many %N you have in the body). So when you write:

#(%)

它翻译成：

(fn [arg1] (arg1))

请注意，这不同于您的第一个匿名函数，如下所示：

Notice that this is different from your first anonymous function, which is like:

(fn [arg1] arg1)

您的版本会将arg1返回一个值，来自展开速记的版本会将其称为函数。您会收到错误，因为字符串不是有效的函数。

Your version returns arg1 as a value, the version that comes from expanding the shorthand tries to call it as a function. You get an error because a string is not a valid function.

由于缩写在主体周围提供了一组括号，因此它只能用于执行单个函数呼叫或特殊形式。

Since the shorthand supplies a set of parentheses around the body, it can only be used to execute a single function call or special form.

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