如何从Lua事件处理函数中获取“自我”？

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
functin 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

在调用函数时，在无法传递self的情况下，是否可以使用self？

如果没有，我应该如何设计？

---

[已解决]我使用翻译，但我想保持礼貌。谢谢您的答复。

Answer 1

避免不必要工作的一种方法是编写一个函数，该函数在将其注册为处理程序之前将为您包装该函数：

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

仅使用调用实际函数的匿名函数

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