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Predicate setarg/3
Availability:built-in
setarg(+Arg, +Term, +Value)
Extra-logical predicate. Assigns the Arg-th argument of the compound term Term with the given Value. The assignment is undone if backtracking brings the state back into a position before the setarg/3 call. If the designated argument of Term is a variable, this variable is unified with Value using normal unification, i.e., setarg/3 behaves as arg/3 in this case. Note that this may produce a cyclic term if Value contains this variable. See also nb_setarg/3.

This predicate may be used for destructive assignment to terms, using them as an extra-logical storage bin. Always try hard to avoid the use of setarg/3 as it is not supported by many Prolog systems and one has to be very careful about unexpected copying as well as unexpected noncopying of terms. A good practice to improve somewhat on this situation is to make sure that terms whose arguments are subject to setarg/3 have one unused and unshared variable in addition to the used arguments. This variable avoids unwanted sharing in, e.g., copy_term/2, and causes the term to be considered as non-ground. An alternative is to use put_attr/3 to attach information to attributed variables (see section 8.1).

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LogicalCaptain said (2020-07-27T15:12:03):0 upvotes 0 0 downvotes
Picture of user LogicalCaptain.

The specification should probably be

setarg(+Arg, !Term, +Value)

Position is 1-based and the predicate fails if Value is out-of-bounds:

?- X=f(a,b,c,d), setarg(0,X,foo).
false.

?- X=f(a,b,c,d), setarg(1,X,foo).
X = f(foo, b, c, d).

?- X=f(a,b,c,d), setarg(4,X,foo).
X = f(a, b, c, foo).

?- X=f(a,b,c,d), setarg(5,X,foo).
false.

Otherwise, works as expected:

?- X=f(a,Y,Y,d), setarg(2,X,Y).
X = f(a, Y, Y, d).
?- X=f(a,Y,Y,d), setarg(2,X,f(Y)).
X = f(a, f(Y), Y, d).

We need more information about unexpected copying as well as unexpected noncopying of terms.