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Oop 为什么超类:initialize()与lua中的SuperClass.initialize(self)不同?_Oop_Lua - Fatal编程技术网

Oop 为什么超类:initialize()与lua中的SuperClass.initialize(self)不同?

Oop 为什么超类:initialize()与lua中的SuperClass.initialize(self)不同?,oop,lua,Oop,Lua,使用时,我遇到了一个奇怪的问题: ParentClass = class('ParentClass') function ParentClass:initialize(param) self.param = param end 现在这将起作用: Child1 = class('Child1', ParentClass) function Child1:initialize(param) ParentClass.initialize(self, param) end 但这不会:

使用时,我遇到了一个奇怪的问题:

ParentClass = class('ParentClass') 

function ParentClass:initialize(param)
   self.param = param
end
现在这将起作用:

Child1 = class('Child1', ParentClass)

function Child1:initialize(param)
  ParentClass.initialize(self, param)
end
但这不会:

Child2 = class('Child2', ParentClass)

function Child2:initialize(param)
  ParentClass:initialize(param)
end  
创建像Child2这样的子类将导致对
param
的修改,而不是针对单个实例,而是针对从父类继承的所有子类

我花了很长时间才发现这个问题,因为我确信
Object:function()
只是
Object.function(Object)
的语法糖,因此应该是相同的。这似乎不是中产阶级的问题,因为我以前使用的另一个OO库也显示出类似的问题

我错过了什么

function Child1:initialize(param)
  ParentClass.initialize(self, param)
end
相当于

function Child1:initialize(param)
  ParentClass.initialize(Child1, param)
end
function Child2:initialize(param)
  ParentClass.initialize(ParentClass, param)
end
所以
self
in

function ParentClass:initialize(param)
   self.param = param
end
function ParentClass:initialize(param)
   self.param = param
end
将参考
Child1

相当于

function Child1:initialize(param)
  ParentClass.initialize(Child1, param)
end
function Child2:initialize(param)
  ParentClass.initialize(ParentClass, param)
end
所以
self
in

function ParentClass:initialize(param)
   self.param = param
end
function ParentClass:initialize(param)
   self.param = param
end
将参考
ParentClass


因此,代码段1将初始化
Child1
,而代码段2将初始化基类。

你说得对。很明显,但我错过了很多天。非常感谢你!