如何从Lua事件处理程序函数获取'self'？

Question

local tbl = {a = 1}
-- Passed and executed function
function tbl:handleFunc(x, y)
print(self.a + x + y)
end
-- I know the above code is syntax sugar.
-- tbl["handleFunc"] = function(self, x, y)
--     print(self.a + x + y)
-- end
-- Register event handlers that I can't control the call
Frame:SetScript("OnClick", tbl.handleFunc)
-- Probably called when an event occurs.
tbl.handleFunc(2, 3)
-- Then it will be like this.
tbl.handleFunc(2, 3, nil)
-- So I wrote a function like this
function tbl.handleFunc(x, y)
local self = tbl  -- This variable is too cumbersome
-- And this way, every time I call a function, I need to check whether the function definition is a dot (.) Or colon (:)
end

当调用函数时，在无法传递self的情况下，有没有方法使用self？

如果没有，我应该如何设计？

---

[Solved]我用了一个翻译，但我想礼貌一点。谢谢你的回答。

Answer 1

避免不必要工作的一种方法是在将函数注册为处理程序之前为您编写一个函数：

local tbl = {a = 1}
function tbl:handleFunc(x, y)
print(self.a + x + y)
end
function wrap_with(obj, func)
return function(...)
return func(obj, ...)
end
end
Frame:SetScript("OnClick", wrap_with(tbl, tbl.handleFunc))

Answer 2

只需使用一个匿名函数即可调用实际函数

Frame:SetScript("OnClick", function(x,y) tbl:handleFunc(x, y) end)