Lua Gideros:具有多个参数的调用函数

时间:2014-10-18 22:37:34

标签: lua gideros

在我使用Gideros Studio的游戏中,我有一个具有多个参数的功能。我想在一个参数上调用我的函数,然后在另一个参数上调用我的函数。这可能吗?

这是我的功能:

local function wiggleroom(a,b,c)
    for i = 1,50 do
        if a > b then
            a = a - 1
        elseif a < b then
            a = a + 1
        elseif a == b then
            c = "correct"
        end
    return c
    end
end

我希望ab进行比较,但稍后会在bc上调用该函数。例如:

variable = (wiggleroom(variable, b, c) --if variable was defined earlier
variable2 = (wiggleroom(a, variable2, c)
variable3 = (wiggleroom(a, b, variable3)

我还希望能够将此函数用于多个对象(每次调用两次参数)。

1 个答案:

答案 0 :(得分:2)

如果我正确理解你,你可以考虑使用lua版本的类。如果您不了解它们,您可能需要查看 this

示例:

tab = {}

function tab:func(a, b, c)  -- c doesn't get used?
    if a then self.a = a end
    if a then self.b = b end
    if a then self.c = c end

    for i = 1,50 do
        if self.a > self.b then
            self.a = self.a - 1
        elseif self.a < self.b then
            self.a = self.a + 1
        elseif self.a == self.b then
            self.c = "correct"
        end
    end
    return c                -- not really necessary anymore but i leave it in
end

function tab:new (a,b,c)    --returns a table
    o = {}
    o.a = a
    o.b = b
    o.c = c
    setmetatable(o, self)
    self.__index = self
    return o
end

                            --how to use:
whatever1 = tab:new(1, 60)  --set a and b
whatever2 = tab:new()       --you also can set c here if needed later in the function

whatever1:func()            --calling your function
whatever2:func(0,64)

print(whatever1.a)          -->51
print(whatever2.a)          -->50
print(whatever1.c)          -->nil
whatever1:func()            --calling your function again
whatever2:func()
print(whatever1.a)          -->60
print(whatever2.a)          -->64
print(whatever1.c)          -->correct
print(whatever2.c)          -->correct