Variables 如何将空变量传递给Lua中的函数
我试图将空值传递给函数,但失败了。这是我的设置Variables 如何将空变量传递给Lua中的函数,variables,lua,function-parameter,Variables,Lua,Function Parameter,我试图将空值传递给函数,但失败了。这是我的设置 function gameBonus.new( x, y, kind, howFast ) -- constructor local newgameBonus = { x = x or 0, y = y or 0, kind = kind or "no kind", howFast = howFast or "no speed" } return setme
function gameBonus.new( x, y, kind, howFast ) -- constructor
local newgameBonus = {
x = x or 0,
y = y or 0,
kind = kind or "no kind",
howFast = howFast or "no speed"
}
return setmetatable( newgameBonus, gameBonus_mt )
end
我只想传递“kind”,并希望构造函数处理其余的内容。喜欢
local dog3 = dog.new("" ,"" , "bonus","" )
或者我只想通过“多快”
我尝试了使用”
和不使用,给出了错误:
“,”附近出现意外符号
nil
是在Lua中表示空的类型和值,因此不应传递空字符串“”
或什么都不传递,而应传递nil
,如下所示:
local dog3 = dog.new(nil ,nil , "bonus", nil )
请注意,可以省略最后一个nil
以第一个参数x
为例,表达式
x = x or 0
相当于:
if not x then x = 0 end
也就是说,如果
x
既不是false
也不是nil
,则将x
设置为默认值0
是我还是调用了错误的函数?您可以调用dog.new
,但函数名为gameBones.new
。
function gameBonus.new( x, y, kind, howFast ) -- constructor
local newgameBonus = type(x) ~= 'table' and
{x=x, y=y, kind=kind, howFast=howFast} or x
newgameBonus.x = newgameBonus.x or 0
newgameBonus.y = newgameBonus.y or 0
newgameBonus.kind = newgameBonus.kind or "no kind"
newgameBonus.howFast = newgameBonus.howFast or "no speed"
return setmetatable( newgameBonus, gameBonus_mt )
end
-- Usage examples
local dog1 = dog.new(nil, nil, "bonus", nil)
local dog2 = dog.new{kind = "bonus"}
local dog3 = dog.new{howFast = "faster"}
function gameBonus.new( x, y, kind, howFast ) -- constructor
local newgameBonus = type(x) ~= 'table' and
{x=x, y=y, kind=kind, howFast=howFast} or x
newgameBonus.x = newgameBonus.x or 0
newgameBonus.y = newgameBonus.y or 0
newgameBonus.kind = newgameBonus.kind or "no kind"
newgameBonus.howFast = newgameBonus.howFast or "no speed"
return setmetatable( newgameBonus, gameBonus_mt )
end
-- Usage examples
local dog1 = dog.new(nil, nil, "bonus", nil)
local dog2 = dog.new{kind = "bonus"}
local dog3 = dog.new{howFast = "faster"}