在序言中将列表的整数元素转换为十进制 [英] Converting integer elements of list to a decimal in prolog

问题描述

我想将一个整数列表转换为一个十进制数，但遇到一个我无法修复的实例化错误.下面是我尝试执行此操作的代码.

I would like to convert an integer list into a decimal number but am getting an instantiation error I cannot fix.Below is my code to try to this.

number([X|[]],X).
number([X|XS],Y) :-
   len([X|XS],B),
   BB is B-1,
   YY is X*10^(BB),
   L is Y+YY,
   number(XS,L).

我不太确定如何解决这个问题.len 是一个返回给定列表长度的函数.任何帮助表示赞赏

I am not too sure how to go over this problem. len is a function to return the length of the given list. Any help is appreciated

推荐答案

假设您有一个十进制数字的整数列表(例如，域为 0–9).你应该能够做这样的事情:

Assuming you have a list of integers that are decimal digits (e.g, having the domain of 0–9). You should be able to do something like this:

digits_value( Ds , V ) :-
  digits_value(Ds,0,V)
  .

digits_value( []     , V , V ) .
digits_value( [D|Ds] , T , V ) :-
  T1 is 10*T + D ,
  digits_value(Ds,T1,V)
  .

尾递归的好处是你可以边走边构建结果.在每次调用时，您只需按 10 倍进行缩放并相加.你不在乎 10 的幂.

The nice thing about the tail recursion is that you build the result as you go. On each call, you just scale by a factor of 10 and add. You don't care about powers of 10.

要按照您尝试的方式进行操作，您可以尝试颠倒列表，以免对 10 的幂大惊小怪:

To do it the way you're trying to do it, you might try reversing the list so as not to fuss with powers of 10:

digits_value(Ds,V) :-
  reverse(Ds,X) ,
  digits_value_x(X,V)
  .

digits_value_x([D],D).
digits_value_x([D|Ds],V) :-
  digits_value_x(Ds,T) ,
  V is 10*T + D
  .

或者，您可以执行以下操作:

Alternatively you could do something like this:

digits_value(Ds,V) :-
  length(Ds,L) ,
  S is 10**(L-1) ,
  digits_value_x(X,S,V)
  .

digits_value_x([D],1,D).
digits_value_x([D|Ds],S,V) :-
  S1 is S / 10 ,
  digits_value_x(Ds,S1,T) ,
  V is D*S + T
  .

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