Question

是否可以在不使用作业的情况下编写此代码摘录？

self.name = self.name.to_s.squeeze(' ').strip

我尝试过使用bang版本的方法，但是如果操作没有执行任何更改（而不是返回nil），则返回self，因此无法使用。

Answer 1

你必须挖掘整个事情。所以它会是这样的：

str.tap {|x| x.squeeze!(' ')}.tap(&:strip!)

这不是我通常建议做的事情。即使你迫切需要变异，最好的代码清晰度来自于按照设计方式使用方法：

str.squeeze!(' ')
str.strip!

如果这样做不方便，请考虑您是否真的需要突变。

Answer 2

如果确实想要避免分配，可以执行此操作：

self.name = 'This  is a    test  '

[['squeeze!', ' '], 'strip!'].each { |cmd| self.name.send(*cmd) }

self.name
# => "This is a test"

Answer 3

对于没有（Ruby）归因且没有tap的单行：

a.name && (a.name.squeeze!(' ') || a.name).strip!

e.g：

$ irb
2.1.1 :001 > class A
2.1.1 :002?>   def name=(name)
2.1.1 :003?>     puts "setting name=#{name.inspect}"
2.1.1 :004?>     @name = name
2.1.1 :005?>     end
2.1.1 :006?>   attr_reader :name
2.1.1 :007?>   end
 => nil 
2.1.1 :008 > a = A.new
 => #<A:0x007fdc909d6df8> 
2.1.1 :009 > a.name = '  and   she    was  '
setting name="  and   she    was  "
 => "  and   she    was  " 
2.1.1 :010 > a.name && (a.name.squeeze!(' ') || a.name).strip!
 => "and she was" 
2.1.1 :011 > a.name
 => "and she was" 
2.1.1 :012 > a.name = 'and   she    was'
setting name="and   she    was"
 => "and   she    was" 
2.1.1 :013 > a.name && (a.name.squeeze!(' ') || a.name).strip!
 => nil 
2.1.1 :014 > a.name = 'and she was '
setting name="and she was "
 => "and she was " 
2.1.1 :015 > a.name && (a.name.squeeze!(' ') || a.name).strip!
 => "and she was"

红宝石与爆炸链接

3 个答案: