Lua - 执行存储在表中的函数

我能够将函数存储到表中。但现在我不知道如何调用它们。最终表将有大约 100 个调用，因此如果可能的话，我想像在 foreach 循环中一样调用它们。谢谢！

该表的定义方式如下：

game_level_hints = game_level_hints or {}
game_level_hints.levels = {}
game_level_hints.levels["level0"] = function()
    return
    {
        [on_scene("scene0")] =
        {
            talk("hint0"),
            talk("hint1"),
            talk("hint2")
        },
        [on_scene("scene1")] =
        {
            talk("hint0"),
            talk("hint1"),
            talk("hint2")
        }
    }
end

Aa 和函数定义：

function on_scene(sceneId)
    -- some code
    return sceneId
end

function talk(areaId)
    -- some code
    return areaId
end

EDIT:

我修改了这些函数，以便它们有更多的上下文。基本上，他们现在返回字符串。我希望发生的是，在调用函数结束时，我将有一个包含所有这些字符串的表（最好是级别表）。

简短的回答：要调用存储在数组中的函数（引用），您只需添加(parameters)，就像你通常做的那样：

local function func(a,b,c) return a,b,c end
local a = {myfunc = func}
print(a.myfunc(3,4,5)) -- prints 3,4,5

事实上，你可以将其简化为

local a = {myfunc = function(a,b,c) return a,b,c end}
print(a.myfunc(3,4,5)) -- prints 3,4,5

长答案：你没有描述你的预期结果是什么，但你写的很可能not做你期望它做的事情。拿这个片段：

game_level_hints.levels["level0"] = function()
    return
    {
        [on_scene("scene0")] =
        {
            talk("hint0"),
        }
    }
end

[This paragraph no longer applies after the question has been updated] You reference on_scene and talk functions, but you don't "store" those functions in the table (since you explicitly referenced them in your question, I presume the question is about these functions). You actually call these functions and store the values they return (they both return nil), so when this fragment is executed, you get "table index is nil" error as you are trying to store nil using nil as the index.

如果你想调用你存储的函数game_level_hints.levels["level0"]，你就这么做game_level_hints.levels["level0"]()