Prolog语法和谓词逻辑

时间:2015-01-09 13:30:13

标签: prolog

我有2个例外点( THIS-RTE )和(其他RTE )我无法设法修复它可以任意1给我一个关于什么是暗示的提示我做错了,因为我很确定我搞错了。感谢

decomp([],[],0,0).
decomp([E|T],R,P,I):-
    0 is mod(E,2),
    decomp(T,R1,P1,I),
    R is [E|R1],    %**(*THIS-RTE*)** ERROR: is/2: Type error: `[]' expected, found `[2|_G7167]' ("x" must hold one character)
    P is 1 + P1.
decomp([E|T],R,P,I):-
    not(0 is mod(E,2)),
    decomp(T,R1,P,I1),
    R is append(E,R1,R), %**(*Other-RTE*)** ERROR: d:/../../../../lab1a.pro:47: evaluable `append(_G5550,_G5551,_G5552)' does not exist, but my clause is right ther
    I is 1 + I1.

append(E,[],[E]).
append(E,[A|St],[A|Et]):-
    append(E,St,Et).

1 个答案:

答案 0 :(得分:3)

你的语法有点不对:

  • R is append(E,R1,R)应该只是append(E,R1,R),因为您正在传递R
  • R is [E|R1]应直接进入规则标题

结果应如下所示:

decomp([],[],0,0).
decomp([E|T], [E|R1], P, I):-
    0 is E mod 2,
    decomp(T,R1,P1,I),
    P is 1 + P1.
decomp([E|T],R,P,I):-
    1 is E mod 2,
    decomp(T,R1,P,I1),
    append(E,R1,R),
    I is 1 + I1.

Demo.