Function 函数中的变量不会在每次调用时重置（LUA）

我正在尝试设置一个函数，将min添加到当前时间对象并返回一个新的时间对象。我为此创建了一个函数，但由于某些原因，每次调用该函数时，函数中的变量都没有被重新设置/本地。 对函数的每次调用都将使用函数中本地变量的过去值，为什么

  local function AddTime (MinAfter, BaseTime) 


        if (MinAfter == nil) then  MinAfter = 0 end
        if (BaseTime == nil) or (BaseTime.min == nil) or (BaseTime.hour == nil)  then  BaseTime = os.date("*t") end

          BaseTime.hour = BaseTime.hour + math.floor((BaseTime.min + MinAfter)/60)
          BaseTime.min =  BaseTime.min + MinAfter - (60 * (math.floor((BaseTime.min + MinAfter)/60)))

          if BaseTime.hour > 24 then  BaseTime.hour = 24 end

          return  BaseTime

        end


        local sunriseHour = os.date("*t" ,os.time {year = 2014, month = 4, day = 19, yday = 259, wday = 4, hour = 6, min = 0, sec = 0, isdst = false});

    -- this is the original time object in this case sunraiseHour

          print ("sunriseHour time:" .. (string.format("%02d",sunriseHour.hour) .. ":" .. string.format("%02d", sunriseHour.min)));


    -- first call 

    local newtime1= AddTime(10, sunriseHour);


          print ("call 1 time:" .. string.format("%02d", newtime1.hour) .. ":" .. string.format("%02d", newtime1.min));

         -- on the 1st call  I get 07:10 which is right 

    -- 2nd call 

    local newtime2= AddTime(10, sunriseHour);


          print ("call 1 time:" .. string.format("%02d", newtime2.hour) .. ":" .. string.format("%02d", newtime2.min));


          -- on the 2nd call  I get 07:20 and not 07:10 since this was the 2nd call to the function - the BaseTime var within the function was not local   

---

当您将

sunrisehoor

传递到AddTime时，它是通过引用而不是通过值传递的，这意味着在AddTime内部对

BaseTime

所做的任何更改都是对

sunrisehoor

的更改——这两个变量（

sunrisehoor

和

BaseTime

）都指向同一个对象

因此，当您在AddTime中编写以下内容时：

BaseTime.hour = BaseTime.hour + math.floor((BaseTime.min + MinAfter)/60)
BaseTime.min =  BaseTime.min + MinAfter - (60 * (math.floor((BaseTime.min + MinAfter)/60)))

您正在修改

日出时间

您似乎不太理解这一点，因为您还在AddTime中为BaseTime指定了一个新值，这表明您认为您有一个新对象。如果要创建sunriseHour的修改副本，则需要在AddTime中执行此操作，或者为时间对象创建某种副本构造函数。

， 我知道我需要复制我的时间对象，否则因为我使用了它的引用，它将修改原始对象。 我找到了一个复制函数，并使用它将对象复制到函数中的本地对象

谢谢

function table.copy(orig)
    local orig_type = type(orig)
    local copy
    if orig_type == 'table' then
        copy = {}
        for orig_key, orig_value in pairs(orig) do
            copy[orig_key] = orig_value
        end
    else -- number, string, boolean, etc
        copy = orig
    end
    return copy
end

 function AddTime (MinAdd, TimeObj) 

local BaseTime = {};
local MinAfter = 0;

if (TimeObj == nil) or (TimeObj.min == nil) or (TimeObj.hour == nil) then  BaseTime =  table.copy(os.date("*t")) else  BaseTime = table.copy(TimeObj)  end;
if (MinAdd == nil) then  MinAfter = 0 else MinAfter = MinAdd end;

    BaseTime.hour = BaseTime.hour + math.floor((BaseTime.min + MinAfter)/60)
    BaseTime.min =  BaseTime.min + MinAfter - (60 * (math.floor((BaseTime.min + MinAfter)/60)))

  if BaseTime.hour > 24 then  BaseTime.hour = 24 end

  return  BaseTime

end






        -- this is the original time object in this case sunraiseHour
         local sunriseHour = os.date("*t" ,os.time {year = 2014, month = 4, day = 19, yday = 259, wday = 4, hour = 6, min = 0, sec = 0, isdst = false});

              print ("sunriseHour time:" .. (string.format("%02d",sunriseHour.hour) .. ":" .. string.format("%02d", sunriseHour.min)));


        -- first call 

        local newtime1= AddTime(10,sunriseHour);


              print ("call 1 time:" .. string.format("%02d", newtime1.hour) .. ":" .. string.format("%02d", newtime1.min));

             -- on the 1st call  I get 07:10 which is right 

        -- 2nd call 

        local newtime2= AddTime(10,sunriseHour);


              print ("call 2 time:" .. string.format("%02d", newtime2.hour) .. ":" .. string.format("%02d", newtime2.min));


              -- on the 2nd call  I get 07:20 and not 07:10 since this was the 2nd call to the function - the BaseTime var within the function become global  




                        print ("Added time:" .. string.format("%02d", AddTime(20, sunriseHour).hour) .. ":" .. string.format("%02d", AddTime(20, sunriseHour).min));

---

在同一个对象上调用AddTime两次，首先添加10，然后添加另一个10，结果是添加20。明白了，但我确实需要保持原始对象不变，因为我想调用AddTime函数，只需保留几个对象。如何创建复制构造函数或创建复制对象的函数。就这样。这和你想的一样简单。我将编辑我的文章以显示一个示例，但我不知道对象可以包含的所有字段。