How to resolve the algorithm Enumerations step by step in the Ruby programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Enumerations step by step in the Ruby programming language
Table of Contents
Problem Statement
Create an enumeration of constants with and without explicit values.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Enumerations step by step in the Ruby programming language
The code defines the two modules Fruits and Card containing constants:
Fruits module:
- Defines constants
APPLE
,BANANA
, andCHERRY
with values0
,1
, and2
, respectively.
Card module:
- SUITS: An array of symbols representing the suits in a deck of cards (
[:Clubs, :Hearts, :Spades, :Diamonds]
). - SUIT_VALUE: A hash created using the
each_with_index
method to associate each suit with its corresponding index. In Ruby 2.1 and later, this is done withSUITS.each_with_index.to_h
. In earlier versions, it would beHash[SUITS.each_with_index.to_a]
. - PIPS: An array of symbols representing the pips (numbers and face cards) in a deck of cards (
[:2, :3, :4, :5, :6, :7, :8, :9, :10, :Jack, :Queen, :King, :Ace]
). - PIP_VALUE: A hash created using
each.with_index(2)
to associate each pip with its corresponding value, starting from 2. Ruby 2.1 and later usesPIPS.each.with_index(2).to_h
, while earlier versions useHash[PIPS.each.with_index(2).to_a]
.
Usage of #include? method with constant symbols:
FRUITS = [:apple, :banana, :cherry]
val = :banana
FRUITS.include?(val) #=> true
This shows how the include? method can be used to check if a value is present in an array of symbols representing the constants defined in the Fruits module.
Source code in the ruby programming language
module Fruits
APPLE = 0
BANANA = 1
CHERRY = 2
end
# It is possible to use a symbol if the value is unrelated.
FRUITS = [:apple, :banana, :cherry]
val = :banana
FRUITS.include?(val) #=> true
module Card
# constants
SUITS = %i(Clubs Hearts Spades Diamonds)
SUIT_VALUE = SUITS.each_with_index.to_h # version 2.1+
# SUIT_VALUE = Hash[ SUITS.each_with_index.to_a ] # before it
#=> {:Clubs=>0, :Hearts=>1, :Spades=>2, :Diamonds=>3}
PIPS = %i(2 3 4 5 6 7 8 9 10 Jack Queen King Ace)
PIP_VALUE = PIPS.each.with_index(2).to_h # version 2.1+
# PIP_VALUE = Hash[ PIPS.each.with_index(2).to_a ] # before it
#=> {:"2"=>2, :"3"=>3, :"4"=>4, :"5"=>5, :"6"=>6, :"7"=>7, :"8"=>8, :"9"=>9, :"10"=>10, :Jack=>11, :Queen=>12, :King=>13, :Ace=>14}
end
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