How to catch exception to backtrack(or Delimited Continuations) in Prolog?

问题

I'm new to Prolog and trying to know some example for catch/3.

The code I want to run is here. (Using swi-prolog v8.0.3)

mylib.pl

mylib([1,2]).
mylib(1).
test(X):- mylib(X), length([1],X).

then consult it in swipl then run the following command.

catch(test(X),error(Err,_Context),format('catch')).

So here I want to catch the typeError exception and get X = 1. But the result is catching the exception and nothing continues. So what should I do to reach my target?

Running messages here.

catch
Err = type_error(integer, [1, 2]),
_Context = context(length/2, _12488).

UPDATE1: I have tried to modify the mylib.pl to the following:

mylib([1,2]).
mylib(1).
test(X):- mylib(X),catch(length([1],X),error(Err,_Context),format('catch')).

Then run test(X). command and got following result.

?- test(X).
catch
X = [1, 2] ;
X = 1.

But what I want is skipping the X = [1, 2]

UPDATE2: got this from reddit I'm exploring the usage of reset/3 and shift/1. It will be great if someone could give an example.

回答1:

Use Prolog's inbuilt failure and backtracking:

mylib([1,2]).
mylib(1). 
test(X):- 
    mylib(X),
    integer(X), % will fail on list, backtrack until a passing value for X is found, i.e. 1
    length([1],X).

回答2:

Only a few Prolog systems provide shift/reset, mostly those that use it for recursive tabling. But if your Prolog system does not have shift/reset, you might help yourself with a meta-interpreter:

solve([], ok).
solve([shift|Y], S) :- !, S = stop(Y).
solve([reset(X,Y)|Z], S) :- !, solve(X, Y), solve(Z, S).
solve([X|Y], S) :- rule(X, L, Y), solve(L, S).

The above returns ok or stop(continuation), depending whether a shift/0 happened or not. The continuation is delimited with reset/2. To have a throw ball, extend it to shift/1 and reset/3.