Prolog - 参数没有充分实例化 [英] Prolog - Arguments are not sufficiently instantiated

问题描述

我正在编写一个小程序，它计算列表中有多少元素不是数字.这是我的代码:

I am writing a little program which counts how many elements in a list are not numbers. Here is my code:

not_number([],0).
not_number([X|T],R):- 
    not(number(X)),
    R1 is R+1,  
    not_number(T,R1). 

not_number([_|Tail],Result):-
    not_number(Tail,Result).  

如果我执行这样的代码:

If I execute code like this :

?- not_number([1,2,3,5], R).

我得到 R = 0(应该是这样)

I am getting that R = 0 (as it should be)

R = 0.

但是如果我把一个字符放在列表中:

But if I put a character in the list:

?- not_number([1,2,3,5,a], R).

然后我收到此错误:

ERROR: not_number/2: Arguments are not sufficiently instantiated
   Exception: (10) not_number([a], _G247) ? 

有人能解释一下代码有什么问题吗?我是 prolog 的新手.

Can someone explain whats wrong with code? I am new to prolog.

推荐答案

我正在写这个答案，因为最好的答案在 lurker 的论点未充分实例化#comment36639272_23815952">评论.我想让它显示为一个实际的答案.

I am writing this answer, because the best answer yet was in a comment by lurker. I want to have it show up as an actual answer.

您的代码不起作用，因为在 not_number([X|T]，R).您的递归案例有点倒退.你想这样做:

Your code is not working, because you're doing R1 is R+1 when R isn't instantiated in the case not_number([X|T], R). Your recursive case is strung a little backwards. You want to do this:

not_number([X|T],R):- 
    not(number(X)),
    not_number(T,R1),
    R is R1+1.

现在is的右边在被调用时被实例化了.

Now the right side of the is is instantiated when it is called.

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