Question

我有以下delayed函数，在每个按钮按下后500毫秒被调用。

function M.delayed1()
  for i = 1, #presses do
    if presses[i].num == 1 then
      presses[i].holdMode = true
      presses[i].shouldUntrigger = false
      return
    end
  end
end

function M.delayed2()
  for i = 1, #presses do
    if presses[i].num == 2 then
      presses[i].holdMode = true
      presses[i].shouldUntrigger = false
      return
    end
  end
end

function M.delayed3()
  for i = 1, #presses do
    if presses[i].num == 3 then
      presses[i].holdMode = true
      presses[i].shouldUntrigger = false
      return
    end
  end
end

function M.delayed4()
  for i = 1, #presses do
    if presses[i].num == 4 then
      presses[i].holdMode = true
      presses[i].shouldUntrigger = false
      return
    end
  end
end

实际上，代码要长得多，因为我总共使用了64个按钮。我想知道在这种情况下是否有可能减少Lua中的代码量。

Answer 1

您可以循环设置M的表项，并为closures分配循环索引作为上限值。

local M = {}
for n = 1, 64 do
  M["delayed" .. tostring(n)] = function()
    for i = 1, #presses do
      if presses[i].num == n then
        presses[i].holdMode = true
        presses[i].shouldUntrigger = false
        return
      end
    end
  end
end

Answer 2

@Henri有您要的；但是我不知道为什么首先要创建64个函数副本？它们似乎都遵循相同的逻辑，那么为什么不使用单个功能来接收带有印刷机编号的参数呢？函数定义看起来像...

local function delayed(presses_array, press_num)
    for id, press in ipairs(presses_array) do
        if press.num == press_num then
            press.holdMode = true
            press.shouldUntrigger = false
            break
        end
    end
end

有没有办法消除Lua中的重复功能？

2 个答案: