这个 lua 代码可以改进吗?

能否重构下面的代码,使其更简洁或更清晰?我还附上了下面的图片来帮助说明我的想法。

    local playerAreaPos = {
    {x = playerPos.x, y = playerPos.y - 1, z = playerPos.z}, -- 北
    {x = playerPos.x, y = playerPos.y + 1, z = playerPos.z}, -- 南
    {x = playerPos.x + 1, y = playerPos.y, z = playerPos.z}, -- 东
    {x = playerPos.x - 1, y = playerPos.y, z = playerPos.z}, -- 西

    {x = playerPos.x - 1, y = playerPos.y + 1, z = playerPos.z}, -- 西南
    {x = playerPos.x + 1, y = playerPos.y + 1, z = playerPos.z}, -- 东南
    {x = playerPos.x - 1, y = playerPos.y - 1, z = playerPos.z}, -- 西北
    {x = playerPos.x + 1, y = playerPos.y - 1, z = playerPos.z} -- 东北
}

local posTable = {
    {x = playerPos.x, y = playerPos.y - 2, z = playerPos.z, dir = "NORTH"},
    {x = playerPos.x, y = playerPos.y - 3, z = playerPos.z, dir = "NORTH"},
    {x = playerPos.x, y = playerPos.y + 2, z = playerPos.z, dir = "SOUTH"},
    {x = playerPos.x, y = playerPos.y + 3, z = playerPos.z, dir = "SOUTH"},

    {x = playerPos.x + 2, y = playerPos.y, z = playerPos.z, dir = "EAST"},
    {x = playerPos.x + 3, y = playerPos.y, z = playerPos.z, dir = "EAST"},
    {x = playerPos.x - 2, y = playerPos.y, z = playerPos.z, dir = "WEST"},
    {x = playerPos.x - 3, y = playerPos.y, z = playerPos.z, dir = "WEST"},

    {x = playerPos.x - 2, y = playerPos.y - 2, z = playerPos.z, dir = "NORTH_WEST"},
    {x = playerPos.x - 3, y = playerPos.y - 3, z = playerPos.z, dir = "NORTH_WEST"},
    {x = playerPos.x + 2, y = playerPos.y - 2, z = playerPos.z, dir = "NORTH_EAST"},
    {x = playerPos.x + 3, y = playerPos.y - 3, z = playerPos.z, dir = "NORTH_EAST"},
    {x = playerPos.x - 2, y = playerPos.y + 2, z = playerPos.z, dir = "SOUTH_WEST"},
    {x = playerPos.x - 3, y = playerPos.y + 3, z = playerPos.z, dir = "SOUTH_WEST"},
    {x = playerPos.x + 2, y = playerPos.y + 2, z = playerPos.z, dir = "SOUTH_EAST"},
    {x = playerPos.x + 3, y = playerPos.y + 3, z = playerPos.z, dir = "SOUTH_EAST"}
}

for i = 1, #posTable do
    if targetPos == Position(posTable[i]) then
        if posTable[i].dir == "NORTH_EAST" then
            print("TELEPORT TO: ", playerAreaPos[8].x, playerAreaPos[8].y)
        elseif posTable[i].dir == "NORTH_WEST" then
            print("TELEPORT TO: ", playerAreaPos[7].x, playerAreaPos[7].y)
        elseif posTable[i].dir == "NORTH" then
            print("TELEPORT TO: ", playerAreaPos[1].x, playerAreaPos[1].y)
        elseif posTable[i].dir == "SOUTH_WEST" then
            print("TELEPORT TO: ", playerAreaPos[5].x, playerAreaPos[5].y)
        elseif posTable[i].dir == "SOUTH_EAST" then
            print("TELEPORT TO: ", playerAreaPos[6].x, playerAreaPos[6].y)
        elseif posTable[i].dir == "SOUTH" then
            print("TELEPORT TO: ", playerAreaPos[2].x, playerAreaPos[2].y)
        elseif posTable[i].dir == "EAST" then
            print("TELEPORT TO: ", playerAreaPos[3].x, playerAreaPos[3].y)
        elseif posTable[i].dir == "WEST" then
            print("TELEPORT TO: ", playerAreaPos[4].x, playerAreaPos[4].y)
        end
    end
end

该函数的目的是将 posTable 中的敌人传送到 playerAreaPos,同时确保它们在相应的行内传送,这意味着如果它们距离主人公向北 3 格,则会被传送到主人公向北 1 格。

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用户6834680
用户6834680
local enemyPos = {x = 11, y = 22, z = 33}
local playerPos = {x = 10, y = 20, z = 30}
local beam_length = 3

if enemyPos.z == playerPos.z then
   local dx = enemyPos.x - playerPos.x
   local dy = enemyPos.y - playerPos.y
   local ax, ay = math.abs(dx), math.abs(dy)
   local len_max, len_min = math.max(ax, ay), math.min(ax, ay)
   if len_max >= 2 and len_max <= beam_length and len_min % len_max == 0 then
      print("TELEPORT TO: ", playerPos.x + dx / len_max, playerPos.y + dy / len_max)
   end
end

本段代码定义了敌人(enemyPos)和玩家(playerPos)的位置变量,并定义了射线长度(beam_length)。

如果敌人和玩家在同一层(z轴相等),则进行以下计算:计算敌人与玩家在x轴上的距离(dx)和在y轴上的距离(dy),在取它们的绝对值后求出最大长度(ax)和最小长度(ay)。

然后,计算出最大和最小长度(len_max和len_min)。如果满足最大长度大于等于2且小于等于射线长度(beam_length),且最小长度可以整除最大长度,则输出“TELEPORT TO: ”和玩家最终的位置(x和y坐标)。

2018-06-07 16:42:36
用户9866482
用户9866482

有许多种方法来实现你的问题,但我打算使用一些数学方法,然后再对它们进行一些考虑。

看一下下面的图表,我们的英雄和敌人之间有一个最小距离,当敌人落入范围内时,它将被传送到英雄的半径限制内,之后,敌人只有在英雄将其扔得远离半径限制时才能传送。

考虑到这些问题,我做了这个实现:

-- returns math functions as local functions
local deg = math.deg
local sin = math.sin
local cos = math.cos
local atan2 = math.atan2

-- returns the distance between two points
local lengthOf = function ( dots )
    local dx, dy = dots.x[2]-dots.x[1], dots.y[2]-dots.y[1]
    return (dx*dx + dy*dy)^.5
end

-- returns the degrees between two points
-- note: 0 degrees is 'east'
local angleBetweenPoints = function ( dots )
    local x, y = dots.x[2]-dots.x[1], dots.y[2]-dots.y[1]
    local radian = atan2(x, y)
    local angle = deg(radian)
    angle = angle < 0 and (360 + angle) or angle

    return angle
end

--设置你的英雄
local hero = {}
hero.posX, hero.posY = 0, 0
hero.radius = 10

--设置敌人
local foe = {}
foe.posX, foe.posY = 0, 18
foe.radius = 8
foe.theta = 0
foe.teleported = false
foe.distToHero = 18
foe.points = {}
foe.curDist = 0

--这部分将是像框架一样的东西
foe.points = {x={hero.posX,foe.posX}, y={hero.posY,foe.posY}}
foe.curDist = lengthOf( foe.points )
if foe.distToHero<=foe.curDist and not foe.teleported then
  foe.theta = angleBetweenPoints ( foe.points )
  foe.posX = hero.radius*sin( foe.theta ) -- 是的,它是反的
  foe.posY = hero.radius*cos( foe.theta ) -- 是的,它是反的
  foe.teleported = true
else
  foe.teleported = false -- 如果敌人被远离英雄击中
end

print(foe.posX,foe.posY)

--打印输出:0, 10

上述实现在地形上放置许多敌人需要大量计算,我的建议是使用C API 或 Box2D,Box2D使用传感器夹具来发生碰撞,并且对这种情况非常有用。当前有许多具有这些功能的Lua SDK,因此您可以快速启动。

2018-06-07 16:43:10