Corona SDK是否足够快以适合矮人堡垒之类的游戏?

时间:2019-03-21 11:19:08

标签: enums lua corona

我想做一个像游戏一样的矮人要塞。 Corona SDK是否足够快?我将如何存储每个单独块的数据。由于我无法在Lua中进行枚举,我该如何定义块的类型?

2 个答案:

答案 0 :(得分:1)

为什么不能在Lua中进行枚举?只需使用一个简单的表即可。

brickType = { A = 1,
              B = 2,
              C = 3,
}

myFristBrick = { weight = 500,
                 volume = 50,
                 type = brickType.C
}

您可以使用Lua表对非常复杂的数据结构进行建模。因此,存储您的实体数据是您最不应该关注的问题。

答案 1 :(得分:1)

我在元表中使用它来模拟enumTypes

local enumType_MT = {
    __newindex = function (t, k, v)
        if t [k] then
            print ('redeclaring enum type ' .. k .. ' not permited')
        else
            rawset (t, k, {})
            local c
            for c = 1, #v do
              rawset (t [k], v[c], c)
              rawset (t[k], c, v[c])
            end
         end
    end
}

local myEnumTypes = {}
setmetatable( myEnumTypes, enumType_MT )

myEnumTypes ['foo']  = {'brick_A', 'brick_B', 'brick_C'}
myEnumTypes ['doo']  = {'brick_E', 'brick_F', 'brick_G'}

myVar = myEnumTypes['foo'].brick_A
print (myVar) -- outputs 1
myVar = myEnumTypes['foo']['brick_C']
print (myVar) -- outputs 3
myVar = myEnumTypes['foo'][2]
print (myVar) -- outputs 'brick_B'

myVar = myEnumTypes['doo'].brick_A
print (myVar) -- outputs nil as brick_A not in doo
myVar = myEnumTypes['doo']['brick_C']
print (myVar) -- outputs nil as brick_C not in doo
myVar = myEnumTypes['doo'][2]
print (myVar) -- outputs 'brick_F'