How to resolve the algorithm Arrays step by step in the Prolog programming language

Published on 12 May 2024 09:40 PM
#Prolog

How to resolve the algorithm Arrays step by step in the Prolog programming language

Table of Contents

Problem Statement

This task is about arrays. For hashes or associative arrays, please see Creating an Associative Array. For a definition and in-depth discussion of what an array is, see Array.

Show basic array syntax in your language. Basically, create an array, assign a value to it, and retrieve an element   (if available, show both fixed-length arrays and dynamic arrays, pushing a value into it). Please discuss at Village Pump:   Arrays.
Please merge code in from these obsolete tasks:

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Arrays step by step in the Prolog programming language

Source code in the prolog programming language

singleassignment:-                   
    functor(Array,array,100), % create a term with 100 free Variables as arguments
                              % index of arguments start at 1
    arg(1 ,Array,a),          % put an a at position 1 
    arg(12,Array,b),          % put an b at position 12
    arg(1 ,Array,Value1),     % get the value at position 1
    print(Value1),nl,         % will print Value1 and therefore a followed by a newline 
    arg(4 ,Array,Value2),     % get the value at position 4 which is a free Variable
    print(Value2),nl.         % will print that it is a free Variable followed by a newline

destructive:-                   
    functor(Array,array,100), % create a term with 100 free Variables as arguments
                              % index of arguments start at 1
    setarg(1 ,Array,a),       % put an a at position 1 
    setarg(12,Array,b),       % put an b at position 12
    setarg(1, Array,c),       % overwrite value at position 1 with c
    arg(1 ,Array,Value1),     % get the value at position 1
    print(Value1),nl.         % will print Value1 and therefore c followed by a newline

listvariant:-
    length(List,100),          % create a list of length 100
    nth1(1 ,List,a),           % put an a at position 1 , nth1/3 uses indexing from 1, nth0/3 from 0
    nth1(12,List,b),           % put an b at position 3
    append(List,[d],List2),    % append an d at the end , List2 has 101 elements
    length(Add,10),            % create a new list of length 10
    append(List2,Add,List3),   % append 10 free variables to List2 , List3 now has 111 elements
    nth1(1 ,List3,Value),      % get the value at position 1
    print(Value),nl.           % will print out a

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