11.3.7 append/3

Synopsis

append(?List1, ?List2, ?List3)

Arguments

List1

list of term

List2

list of term

List3

list of term

A list consisting of List1 followed by List2.

Description

Appends lists List1 and List2 to form List3:

| ?- append([a,b], [a,d], X).

X = [a,b,a,d]

| ?- append([a], [a], [a]).

no
| ?- append(2, [a], X).

no

Takes List3 apart:

| ?- append(X, [e], [b,e,e]).

X = [b,e]

| ?- append([b|X], [e,r], [b,o,r,e,r]).

X = [o,r]

| ?- append(X, Y, [h,i]).

X = [], 
Y = [h,i] ;

X = [h], 
Y = [i] ;

X = [h,i], 
Y = [] ;

no

Backtracking

Suppose L is bound to a proper list. That is, it has the form [T1,…,Tn] for some n. In that instance, the following things apply:

1. append(L, X, Y) has at most one solution, whatever X and Y are, and cannot backtrack at all.
2. append(X, Y, L) has at most n+1 solutions, whatever X and Y are, and though it can backtrack over these it cannot run away without finding a solution.
3. append(X, L, Y), however, can backtrack indefinitely if X and Y are variables.

Examples

The following examples are perfectly ordinary uses of append/3:

To enumerate adjacent pairs of elements from a list:

next_to(X, Y, (*in*) List3) :-
        append(_, [X,Y|_], List3).

To check whether Word1 and Word2 are the same except for a single transposition. (append/5 in library(lists) would be better for this task.)

one_transposition(Word1, Word2) :-
        append(Prefix, [X,Y|Suffix], Word1),
        append(Prefix, [Y,X|Suffix], Word2).

| ?- one_transposition("fred", X).
X = "rfed" ;
X = "ferd" ;
X = "frde" ;
no

Given a list of words and commas, to backtrack through the phrases delimited by commas:

comma_phrase(List3, Phrase) :-
        append(F, [','|Rest], List3),
        !,
        (   Phrase = F
        ;   comma_phrase(Rest, Phrase)
        ).
comma_phrase(List3, List3).

| ?- comma_phrase([this,is,',',um,',',an, example], X).
X = [this,is] ;
X = [um] ;
X = [an,example] ;
no

Exceptions

None.

See Also

ref-lte-acl, library(lists).