Lua metatables and metamethod - How to call a different member function

问题

I have the following Class

local PROGRESS = {}

PROGRESS.__index = function(self,key)

    if key~="__group" and self.__group[key] then 
        return  self.__group[key]           
    else 
        return rawget(self,key)
    end 
end

What this does is when You access table[key] it performs a lookup in table.__group (which is an object of another class) and returns table.__group[key] ,if it is not nil.

Now I am trying to do the same for member functions. i.e If I call table:key() a lookup must be performed in table.__group and if the function is present, then table.__group:key() should be called.

How do I accomplish this?

I tried to do this.

local PROGRESS = {}

    PROGRESS.__index = function(self,key)

       if key~="__group" and self.__group[key] then 


         local val = self.__group[key]
         if type(val) == "function" then 
             self.__group:val()
             return function() end 
         end


         return  self.__group[key]          
       else 
         return rawget(self,key)
       end 
    end

But there are 2 things wrong here.

I am unable to retrieve the original function's arguments
Event if I just ACCESS table[key].function without calling it, the function will be called

And I've got the feeling that I am trying to complicate things and the solution is way simpler.

Any help is appreciated.

UPDATE

@Mud The problem with the original code is that the object passed as 'self' to the member function is an object of the new class. Not of the old class.

Consider this code

GROUP_CLASS = {}
GROUP_CLASS.__index = GROUP_CLASS
function GROUP_CLASS:showSum    (a,b) print(self);print(a + b) end


group_object = setmetatable({},GROUP_CLASS)
group_object:showSum(1,2)





local PROGRESS_CLASS = {}
PROGRESS_CLASS.__index = function(self,key,value)
    if key~="__group" and self.__group[key] then 
        return self.__group[key]
    else 
        return rawget(self,key)
    end 
end 



progress_object = setmetatable( {__group = group_object} , PROGRESS_CLASS)
progress_object:showSum(3,3) 
--progress_object is passed as first argument to showSum.  But i need group_object to be passed

In the above code, When progress_object:showSum(3,3) is called, is it possible to pass group_object (or in other words progress_object.__group) as self instead of progress_object.

Hope that makes sense.

回答1:

Response to updated post:

progress_object is passed as first argument to showSum. But i need group_object to be passed

If you're going to ignore the state of the object a method is called on, and substitute the state of some other object, why is it even a method on that object? That's like overriding the addition operator to do multiplication, a recipe for confusion.

In other words, you want this:

progress_object:method("foo")

To resolve, via bizarre internal machinery, into this:

group_object:method("foo")

Why not skip a step and just make the latter call?

If you must, you could achieve this by returning a proxy for the method which replaces self with __group

local PROGRESS_CLASS = {}
PROGRESS_CLASS.__index = function(self,key)
  local groupval = self.__group[key]
  if key == '__group' or not groupval then
    return rawget(self,key)
  elseif type(groupval) ~= 'function' then
    return groupval
  else
      return function(...)
        if self == ... then -- method call
          -- replace self argument with __group
          return groupval(self.__group,select(2,...))
        else
          return groupval(...)
        end
      end
  end
end

Response to original post:

How I am trying to do the same for member functions. i.e If I calltable:key() a lookup must be performed in table.__group and if the function is present, then table.__group:key() should be called.

How do I accomplish this?

Do nothing. Your original code handles this.

Lua doesn't know what a "member function" is. A member is a member (i.e. an element in a table), and whether the value of that member is a function is irrelevant.

Remember:

obj:method(a,b,c) is exactly equivalent to obj.method(obj,a,b,c)
obj.method is exactly equivalent to obj["method"].
Your code already resolves obj["method"] into obj.__group["method"]

So you're done.

For instance, say we have:

group = {}
function group:showSum    (a,b) print(a + b) end
function group:showProduct(a,b) print(a * b) end

Using your first code, we can write:

foo = setmetatable({__group = group}, PROGRESS)

foo:showSum(3,3) -- 6
foo:showProduct(3,3) -- 9

That's it.

Now, as long as we're here, let's look at what your second function is doing:

     local val = self.__group[key]
     if type(val) == "function" then 
         self.__group:val()
         return function() end 
     end

First you grab the function value from __group. At this point you're done. Simply return that value, and the caller is going to call that value (i.e. (...)). Instead, you call __group["val"] which is likely a totally different function from __group[key] (unless key=="val"), then you pass the caller a function which does nothing.

来源：https://stackoverflow.com/questions/11232680/lua-metatables-and-metamethod-how-to-call-a-different-member-function

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