Question

我在代码中遇到了意外（对我而言）的行为，所以我试图在REPL中找出问题所在。但是，这些构造函数似乎都具有相同的结果（空哈希）：

irb> a = {}
# => {}

irb> b = Hash.new(0)
# => {}

当我将{}传递给reduce函数时，我得到了NoMethodError。这两个构造函数之间有什么区别？

irb> arr = "count the occurance of each of the words".scan(/\w+/)
# => ["count", "the", "occurance", "of", "each", "of", "the", "words"]

irb> x = arr.reduce(Hash.new(0)) { |hsh, word| hsh[word] += 1; hsh }
# => {"count"=>1, "the"=>2, "occurance"=>1, "of"=>2, "each"=>1, "words"=>1}

irb> x = arr.reduce({}) { |hsh, word| hsh[word] += 1; hsh }
# NoMethodError: undefined method `+' for nil:NilClass

Answer 1

Hash.new(0)为0的任意键设置默认值，而{}设置nil

h1 = Hash.new(0)
h1.default  # => 0
h1[:a] += 1 # => 1
h2 = {}
h2.default  # => nil
h2[:a] += 1 # => NoMethodError: undefined method `+' for nil:NilClass

Answer 2

如果我们使用Hash.new（0）创建一个哈希对象，那么在这种情况下，参数0将被用作哈希的默认值 - 如果你查找一个不是的键，它将是返回的值但是在哈希

中

def count_frequency(word_list)
  counts = Hash.new(0)
  word_list.each { |word|
    counts[word] += 1
  }
   counts
end
p count_frequency(%w(red blue green red white black))

将产生

{"red"=>2, "blue"=>1, "green"=>1, "white"=>1, "black"=>1}

添加更多内容

brands= Hash.new( "nike" )
puts "#{brands[0]}"
puts "#{brands[101]}"

两者都会产生

nike
nike

和

brands = {0 => "nike" , 1 =>"Reebok"}
brands[0]   # => "nike"
brands[1]   # => "Reebok"
brands[100] # => nil

＆＃34; Hash.new（0）＆＃34;之间的区别是什么？和＆＃34; {}＆＃34;

2 个答案: