Question

我正在玩一个小小的地址簿程序。我有三节课; AddressBook，Contact和Address。在to_s类中编写的Address方法定义了地址的输出格式，我在print_addresses类中编写了一个Contact方法，该方法是否传入了参数是long或short应输出to_s方法中定义的两种输出格式之一。在print_addresses('long')类的Contact实例上调用AddressBook时，它只会输出short方法中定义的to_s格式。

这不符合我的预期，我很确定我对方法和变量范围或显式和隐式返回的理解是错误的，但我无法辨别出错的地方。我写的代码如下，任何人都可以帮助我更好地理解这个吗？谢谢。

地址类

class Address
  attr_accessor :kind, :street_1, :street_2, :city, :state, :postal_code 


  def to_s(format = 'short')
    address = ''
    case format
    when 'long'
      address += street_1 + "\n"
      address += street_2 + "\n" if !street_2.nil? 
      address += "#{city}, #{state} #{postal_code}"
    when 'short'
      address += "#{kind}: "
      address += street_1
      if street_2 
        address += " " + street_2 
      end
      address += ", #{city}, #{state}, #{postal_code}"
    end
    address
   end
  end

联系班级

require './phone_number'
require './address'

class Contact
  attr_writer :first_name, :middle_name, :last_name
  attr_reader :phone_numbers, :addresses

  def initialize
    @phone_numbers = []
    @addresses = []
  end

def add_address(kind, street_1, street_2, city, state, postal_code)
    address = Address.new
    address.kind = kind
    address.street_1 = street_1
    address.street_2 = street_2
    address.city = city
    address.state = state
    address.postal_code = postal_code
    addresses.push(address)
end

def print_addresses(format = 'short')
    puts "Addresses:"
    case format
    when 'short' 
      addresses.each do |address|
        address.to_s('short')
      end
    when 'long'
      addresses.each do |address|
        address.to_s('long')
      end
    end
  end
end

AddressBook类

require './contact'

class AddressBook
  attr_reader :contacts

  def initialize
    @contacts = []
  end

  def find_by_name(name)
    results = []
    search = name.downcase
    @contacts.each do |contact|
      if contact.first_name.downcase.include?(search)
        results.push(contact)
      end
    end
    puts "Name search results (#{search})"
    results.each do |contact|
      puts contact.to_s('full_name')
      puts contact.print_addresses('long')
      puts contact.print_phone_numbers

    end
  end
end

address_book = AddressBook.new

james = Contact.new
james.first_name = "James"
james.middle_name = "jim"
james.last_name = "jimbo"
james.add_address('home', '1', '2', '3', '4', '5')



address_book.contacts.push(james)
puts "-" * 35
puts james.print_addresses('long')
puts '=' * 35

ruby address_book.rb - 输出

------------------------------
Addresses:
home: 1, 2, 3, 4, 5
===================================

期待并试图实现：

------------------------------
Addresses:
1,
2,
3, 4, 5
===================================

Answer 1

它正在做长短版本。您可以在puts "long"函数中添加puts "short"和to_s来查看此内容。

问题不在于范围，而在于您依赖print_addresses中最后一次评估表达式的隐式返回。 print_addresses不会打印任何内容，也不会显式返回任何内容。它尽职地调用address.to_s('long')，抛出结果，并返回最后评估的表达式的结果...... address。然后address运行puts，这会导致它被转换为具有默认短格式的字符串。您可以注释掉address.to_s的来电，您将获得相同的结果。

address是最后一个评估表达式，这可能看起来很奇怪，这就是为什么你应该总是做一个明确的回归。它使代码更易于阅读，并为每个人节省了很多麻烦。这也指出你的函数应该返回多个东西，这是你不能轻易做到的。

  def print_addresses(format = 'short')
    puts "Addresses:"

    formatted_addresses = []

    case format
    when 'short' 
      addresses.each do |address|
        formatted_addresses << address.to_s('short')
      end
    when 'long'
      addresses.each do |address|
        formatted_addresses << address.to_s('long')
      end
    end

    return formatted_addresses
  end

作为旁注，由于print_addresses不打印，因此应将其称为format_addresses。它也应该摆脱冗余的案例陈述。额外格式化（“地址”标题）进入一个包装方法，使format_addresses变得灵活。

  def display_addresses(format)
    return ["Addresses: "] + format_addresses(format)
  end

  def format_addresses(format = 'short')    
    formatted_addresses = []

    addresses.each do |address|
      formatted_addresses << address.to_s(format)
    end

    return formatted_addresses
  end

Ruby explicit＆amp;隐含的回归混淆

1 个答案: