super.是什么意思？在红宝石中做什么？
使用以下代码：
class ObjA
def func
puts "ObjA"
end
end
module Mod
def func
puts "Mod"
end
end
class ObjB < ObjA
include Mod
def func
puts "super called"
super
puts "super.func called"
super.func
end
end
运行 ObjB.new.func 结果：
ruby-1.9.2-p180 :002 > ObjB.new.func
super called
Mod
super.func called
Mod
NoMethodError: undefined method `func' for nil:NilClass
from test.rb:19:in `func'
from (irb):2
我理解 super 的作用 - 它调用超类的当前方法。 include Mod 使 Mod 成为下一个超类，因此调用 Mod#func 。
然而，super.func 正在做什么？我认为它相当于 super，但虽然它确实打印出相同的输出，但它也会抛出 NoMethodError。
With the following code:
class ObjA
def func
puts "ObjA"
end
end
module Mod
def func
puts "Mod"
end
end
class ObjB < ObjA
include Mod
def func
puts "super called"
super
puts "super.func called"
super.func
end
end
Running ObjB.new.func results in:
ruby-1.9.2-p180 :002 > ObjB.new.func
super called
Mod
super.func called
Mod
NoMethodError: undefined method `func' for nil:NilClass
from test.rb:19:in `func'
from (irb):2
I understand what super does - it calls the current method on the superclass. include Mod makes Mod the next superclass so Mod#func is called.
However, what is super.func doing? I thought it would be equivalent to super, but while it does print out the same output, it also throws a NoMethodError.
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我假设 super.func 会做与任何形式的方法链接相同的事情。它调用
super，然后对super返回的结果调用func。super部分将调用Mod#func，打印出“Mod”，然后在Mod# 的返回值上调用，即 nil（这是因为funcfuncputs返回 nil）。由于 nil 没有func方法，所以它说I assume
super.funcwould do the same thing as any form of method chaining. It callssuper, and then callsfuncon the result returned bysuper.The
superpart would callMod#func, which prints out "Mod", then callsfuncon the return value ofMod#func, ie nil (that's becauseputsreturns nil). As nil doesn't have afuncmethod, it says