在Clojure中，如何解构映射的所有键？ [英] In Clojure, how to destructure all the keys of a map?

问题描述

 （let [{：keys [ cpp js]} {：cpp 88：js 90}] 
（println js）; 90 
（println cpp）; 88 
）
  / pre> 
 
 
是否有办法重组所有地图的键？
 
 
 
也许像：
 （let [{：all-the-keys} {：cpp 88：js 90} 
（println js）; 90 
（println cpp）; 88 
）
  
 
 
解决方案
不是真的，这不是一个好主意。想象：
 （let [{：all-the-keys} m] 
（foo bar））
  
是foo还是bar全局变量？当地人？你应该从m中提取的密钥？如果有时包含foo键，此代码应该做什么，foo也是一个全局函数？有时你调用全局的，有时你调用函数存储在m？
 
 
 
忽略技术问题（大部分可能被克服），这真的是一个灾难用于可读性和可预测性。只要明确你想要拉出什么键;如果你经常想拉出相同的十个键，你可以写一个简单的宏，如（with-person p body），这样就简化了这种情况。
 
In clojure, it is possible to destructure some keys of a map like this:
(let [{:keys [cpp js]} {:cpp 88 :js 90}] 
   (println js); 90
   (println cpp); 88
 )
Is there a way to destructure all the keys of a map?

Maybe something like:
(let [{:all-the-keys} {:cpp 88 :js 90}] 
   (println js); 90
   (println cpp); 88
 )

解决方案
Not really, and it wouldn't be a good idea. Imagine:
(let [{:all-the-keys} m]
  (foo bar))
Are foo and bar globals? Locals? Keys you should extract from m? What should this code do if m sometimes contains a foo key, and foo is also a global function? Sometimes you call the global, and sometimes you call the function stored in m?

Ignoring the technical problems (most of which can probably be overcome), it is really a disaster for readability and predictability. Just be explicit about what keys you want to pull out; if you frequently want to pull out the same ten keys, you can write a simple macro like (with-person p body) that simplifies that common case for you.

                        
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