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Ruby,我找不到下面的代码是如何工作的

ruby

0

John_ · 技术社区 · 6 年前

class ArrayMine < Array

  def join( sep = $,, format = "%s" )

    collect do |item|
      sprintf( format, item )
    end.join( sep )
  end

end

=> rooms = ArrayMine[3, 4, 6]    #i couldn't understand how this line works

print "We have " + rooms.join( ", ", "%d bed" ) + " rooms available."

我对String也试过,但它出现了一个错误。

    thank you...

3 回复 | 直到 15 年前

1

3

Jonas Elfström 15 年前

ArrayMine继承自Array,您可以像这样初始化Ruby数组。

>> rooms = Array[3,4,6]
=> [3, 4, 6]

与

>> rooms = [3,4,6]
=> [3, 4, 6]

也很奇怪的样子 def join( sep = $,, format = "%s" ) 正在使用 pre-defined variable $, 这是打印和 Array#join

也可以这样做

rooms=["b",52,"s"]
print "We have " + rooms.map{|s| s.to_s+" bed"}.join(", ") + " rooms available."

你不能做你想做的事的原因 String 是因为赋值不是类方法,而是 Array 是。只是 new

>> s = Substring.new("abcd")
=> "abcd"
>> s.checking_next
=> "abce"

在Ruby中不能重写赋值,这只是因为setter方法看起来像赋值,但实际上它们是方法调用,因此可以重写它们。如果你觉得自己很狡猾,而且还想有类似的行为 a=SubArray[1,2,3] 你可以创建一个 << class method 像这样的:

class Substring < String
  def next_next()
    self.next().next() 
  end
  def self.<<(val)
    self.new(val)
  end
end 

>> sub = Substring<<"abcb"
=> "abcb"
>> sub.next_next
=> "abcd"
>> sub<<" the good old <<"
=> "abcb the good old <<"
>> sub.class<<"this is a new string"
=> "this is a new string"

2

1

YOU 15 年前

%d bed 到 %s bed

irb(main):053:0> rooms = ArrayMine["a","b","c"]
=> ["a", "b", "c"]

irb(main):055:0> print "We have " + rooms.join( ", ", "%s bed" ) + " rooms available."
We have a bed, b bed, c bed rooms available.=> nil

3

1

Josh Lee 15 年前

你只是在调用 []

class Spam
  def self.[]( *args )
    p args
  end
end

>> Spam[3,4,5]
[3, 4, 5]