Prolog列表中元素的总和 [英] Sum of elements in list in Prolog

问题描述

list_sum([], 0).
list_sum([Head | Tail], TotalSum) :-
    list_sum(Tail, Sum1),
    Total = Head + Sum1.

此代码返回true.如果将Total = Head + Sum1替换为Total is Head + Sum1，则它将返回该值.但是我应该将其替换为这样的结果:

This code returns true. If I replace Total = Head + Sum1 with Total is Head + Sum1, then it will return the value. But what I should replace it with to get the result like this:

?- list_sum([1,2,0,3], Sum).
Sum = 1+2+0+3 ; % not to return value 6!!!

推荐答案

请注意，在过程的第二个子句中，永不实例化TotalSum.在查询您的代码时，您应该已经收到解释器的警告.

Note that in the second clause of your procedure TotalSum is never instantiated. You should have received a warning by the interpreter when consulting your code.

这是我的建议:

list_sum([Item], Item).
list_sum([Item1,Item2 | Tail], Total) :-
    list_sum([Item1+Item2|Tail], Total).

第一个子句处理基本情况，当列表中仅剩一个元素时，即为结果.

The first clause deals with the base case, when there is only one element left in the list, that is your result.

第二个子句处理递归步骤.它获取列表的前两个项目，并执行递归调用，用新的术语Item1 + Item2替换这两个项目.

The second clause deals with the recursion step. It grabs the first two items of the list and performs a recursive call replacing those two items with a new term Item1+Item2.

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