how to pass a list in a predicate in prolog

2020-04-26 01:35发布

问题:

I want to store a paragram as a list in a variable and then call that list for counting how many times a particular word appears in that paragraph.

However, when I do this:

L = [hello,hello,hello].

counthowmany(_, [], 0) :- !.
counthowmany(X, [X|Q], N) :- !, counthowmany(X, Q, N1), N is N1+1.
counthowmany(X, [_|Q], N) :- counthowmany(X, Q, N).

... and compile buffer, and then ask this:

counthowmany(hello,L,N). 

The number of "hello" occurrences in the list doesn't show, instead I receive a warning:

singleton variable:[X]

回答1:

The line in a prolog file:

L = [hello,hello,hello].

Means to prolog:

=(L, [hello,hello,hello]).

Which means you're attempting to define a predicate, =/2. So not only will you get a singleton warning about L (since L isn't used anywhere else in this predicate definition), but you'll also see an error about an attempt to re-define the built-in =/2 since prolog already has it defined.

What you can do instead is:

my_list([hello,hello,hello]).

Then later on, you can do:

my_list(L), counthowmany(hello,L,N).

Note that this case works:

L = [hello,hello,hello], counthowmany(hello,L,N).

It works because it's not attempting to re-define =/2. It is just using the existing built-in predicate =/2.



回答2:

You do

?- X = [hello,how,are,you,hello,hello], counthowmany(hello, X, N).
X = [hello, how, are, you, hello, hello],
N = 3.

First you first bind X ans then you ask for this specific X.

Example 2.

?- counthowmany(hello, X, N).
X = [],
N = 0.


标签: prolog