What is the difference between a list with anythin

问题:

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.

Example:

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

R = [1]

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

If I do this:

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).

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).

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

回答1:

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]])

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.

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).

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 : ]

回答2:

_ is anything... foo, [1,2], bar(42,foo[2,3,7]) etc
[_] is a list that has exactly one element that could be anything

in your example if L2 has more than one elements (or is the empty list) then it won't match with [_]