有任何内容([_])和任何内容(_)的列表有什么区别 [英] What is the difference between a list with anything ([_]) and anything (_)

问题描述

我试图完成以下任务，如果我有两个列表L1和L2，我希望结果(R)是从L1减去L2.

I was trying to accomplish the following, if I have two lists, L1 and L2, I wanted that the result (R) to be the "subtraction" of L2 from L1.

示例:

L1 = [1,2,3]
L2 = [2,3,4,5]

R = [1]

我能够做到这一点，但我不知道_和[_]有什么区别.

I WAS able to accomplish this but I can't tell what is the difference between _ and [_].

如果我这样做:

diferencia([],_,[]).
diferencia([X|Tail],L2,R):-
    member(X,L2),
    diferencia(Tail,L2,R).
diferencia([X|Tail],L2,[X|R]):-
    not(member(X,L2)),
    diferencia(Tail,L2,R).

它起作用了，如果我这样做，就会给我错误:

It works, if I do this, it gives me false:

diferencia([],[_],[]).
diferencia([X|Tail],L2,R):-
    member(X,L2),
    diferencia(Tail,L2,R).
diferencia([X|Tail],L2,[X|R]):-
    not(member(X,L2)),
    diferencia(Tail,L2,R).

我认为包含任何[_]内容的列表都应该起作用，因为L2始终是列表.

I would assume a list containing anything [_] should work since L2 will always be a list.

推荐答案

实际上，_仅匹配一个变量和一个变量.在这里，您希望它与2, 3, 4, 5(四个变量)匹配.不可以它只能匹配[2, 3, 4, 5](列表).您必须编写[_|_]以便头部和尾部匹配([2|[3, 4, 5]])

Actually, _ matches one variable and one variable only. Here, you'd want it to match 2, 3, 4, 5 (the four variables). It can't. It can only match [2, 3, 4, 5] (the list). You'd have to write [_|_] so that the head and the tail are matched ([2|[3, 4, 5]])

或[_, _, _, _, _, _, ...]，其中_的数量是列表中项目的确切数量，以便每个元素都与匿名变量正确匹配.

Or [_, _, _, _, _, _, ...] with the number of _ being the exact number of items in your list so that every single element is properly matched with an anonymous variable.

要记住的基本内容是_只是一个普通变量.如果您在记住它时遇到麻烦，只需使用显式名称即可，例如_Head或_Accumulator，这样您就可以在编写代码时意识到，所操作的对象实际上是一个变量，只在乎它(变量开头至少在swi-pl中使用_不会产生单例变量警告，因此可以使用它们代替_以获得更好的整体清晰度.

The basic thing to remember is that _ is just a normal variable. If you have troubles remembering it, just explicit names, like _Head or _Accumulator, so that you do realize when you write your code that the thing you manipulate is actually a variable, only you do not care about it (variable starting with a _ won't produce a singleton variable warning, in swi-pl at least, so they're usable instead of _ for a better overall clarity).

编辑:另一种说法是，标题中您认为_是任何东西.但是，任何事物都可以是虚无的，任何事物都可以是许多事物. _只能是一件事.这就是为什么它不起作用:]

Edit : one other way to say it is that in your title, you think _ is anything. But anything can be nothing, and anything can be many things. _ can only be one thing. That's why it doesn't work : ]

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