在Lisp中将十进制数转换为八进制

问题描述

我正在尝试在Common Lisp中编写一个函数，以递归方式将以10为底的数字转换为以列表表示的以8为底的数字.

I'm trying to write a function in Common Lisp to convert a base 10 number into a base 8 number, represented as a list, recursively.

这是我到目前为止所拥有的:

Here's what I have so far:

(defun base8(n)
(cond
    ((zerop (truncate n 8)) (cons n nil))
    ((t) (cons (mod n 8) (base8 (truncate n 8))))))

当我输入数字<时此功能可以正常工作8和> -8，但是递归的情况给我带来了很多麻烦.当我尝试将8作为参数(应返回(1 0))时，出现错误Undefined operator T in form (T).

This function works fine when I input numbers < 8 and > -8, but the recursive case is giving me a lot of trouble. When I try 8 as an argument (which should return (1 0)), I get an error Undefined operator T in form (T).

提前谢谢.

推荐答案

似乎您已经忘记了(defun t ...)了，或者这不是您打算在cond中使用的功能t吗?也许是t真值?

It seems you have forgotten to (defun t ...) or perhaps it's not the function t you meant to have in the cond? Perhaps it's t the truth value?

Common Lisp的双重名称空间性质使t既可以是函数又可以是真值.区别在于您使用的是哪种上下文，并且您显然正在尝试将t用作函数/宏.

The dual namespace nature of Common Lisp makes it possible for t to both be a function and the truth value. the difference is which context you use it and you clearly are trying to apply t as a function/macro.

这里是为真值编辑的代码，而不是t函数:

Here is the code edited for the truth value instead of the t function:

(defun base8(n)
  (cond
    ((zerop (truncate n 8)) (cons n nil))
    (t (cons (mod n 8) (base8 (truncate n 8))))))

(base8 8) ; ==> (0 1)

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