shovel（<<）运算符在Ruby Hashes中如何工作？ [英] How does shovel (<<) operator work in Ruby Hashes?

问题描述

我正在阅读 Ruby Koans教程系列，当我在 about_hashes.rb


  def test_default_value_is_the_same_object 
 hash = Hash.new（[]）
 
 hash [：one]<< uno
 hash [：two]<< dos
 
 assert_equal [uno，dos]，hash [：one] 
 assert_equal [uno，dos]，hash [：two] 
 assert_equal [uno，dos]，hash [：three] 
 
 assert_equal true，hash [：one] .object_id == hash [：two] .object_id 
 end 
  

assert_equals 中的值实际上是预期的教程。但我无法理解如何使用<< = 运算符？ p>



我的期望是：



• hash [：one ] 会是 [uno]


• hash [：two] 会是 [dos]


• hash [：three] 会是 []

为什么我的期望是错的？

解决方案

当你在做 hash = Hash.new（[] ）您正在创建一个Hash，其默认值与所有键完全相同的Array实例。所以，当你访问一个不存在的键时，你会得到相同的数组。

  h = Hash.new （[]）
h [：foo] .object_id＃=> 12215540 
 h [：bar] .object_id＃=> 12215540 
  

如果你想为每个键一个数组，你必须使用 Hash.new ：

  h = Hash.new {| h，k | h [k] = []} 
 h [：foo] .object_id＃=> 7791280 
 h [：bar] .object_id＃=> 7790760 
  

编辑：另请参阅Gazler关于＃<< 方法以及您实际调用它的对象。

I was going through Ruby Koans tutorial series, when I came upon this in about_hashes.rb:

def test_default_value_is_the_same_object
  hash = Hash.new([])

  hash[:one] << "uno"
  hash[:two] << "dos"

  assert_equal ["uno", "dos"], hash[:one]
  assert_equal ["uno", "dos"], hash[:two]
  assert_equal ["uno", "dos"], hash[:three]

  assert_equal true, hash[:one].object_id == hash[:two].object_id
end

The values in assert_equals, is actually what the tutorial expected. But I couldn't understand how there is a difference between using << operator and = operator?

My expectation was that:

• hash[:one] would be ["uno"]
• hash[:two] would be ["dos"]
• hash[:three] would be []

Can someone please explain why my expectation was wrong?

解决方案

When you're doing hash = Hash.new([]) you are creating a Hash whose default value is the exact same Array instance for all keys. So whenever you are accessing a key that doesn't exist, you get back the very same Array.

h = Hash.new([])
h[:foo].object_id # => 12215540
h[:bar].object_id # => 12215540

If you want one array per key, you have to use the block syntax of Hash.new:

h = Hash.new { |h, k| h[k] = [] }
h[:foo].object_id # => 7791280
h[:bar].object_id # => 7790760

Edit: Also see what Gazler has to say with regard to the #<< method and on what object you are actually calling it.

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