【背景】
泛型 for 在自己内部保存迭代函数，实际上它保存三个值：迭代函数、状态常量、控制变量。
下面我们看看泛型 for 的执行过程：
1.首先，初始化，计算 in 后面表达式的值，表达式应该返回泛型 for 需要的三个值：迭代函数、状态常量、控制变量；与多值赋值一样，如果表达式返回的结果个数不足三个会自动用 nil 补足，多出部分会被忽略。
2.第二，将状态常量和控制变量作为参数调用迭代函数（注意：对于 for 结构来说，状态常量没有用处，仅仅在初始化时获取他的值并传递给迭代函数）。
3.第三，将迭代函数返回的值赋给变量列表。（k, v为变量列表）
4.第四，如果返回的第一个值为nil循环结束，否则执行循环体。
5.第五，回到第二步再次调用迭代函数
在Lua中我们常常使用函数来描述迭代器，每次调用该函数就返回集合的下一个元素。Lua 的迭代器包含以下两种类型：
1.无状态的迭代器
2.多状态的迭代器
迭代器的状态包括被遍历的表（循环过程中不会改变的状态常量）和当前的索引下标（控制变量），ipairs 和迭代函数都很简单，我们在 Lua 中可以这样实现：
function iter (a, i)
i = i + 1
local v = a[i]
if v then --“if v”的意思是“当v为nil或false时该条件语句的判断为false，不执行分支语句”，但事实上，v可以为false，虽然当v为false时并不满足if v 这个条件——看附录。
return i, v
end
end
function ipairs (a)
return iter, a, 0
end
当 Lua 调用 ipairs(a) 开始循环时，他获取三个值：迭代函数 iter、状态常量 a、控制变量初始值 0；然后 Lua 调用 iter(a,0) 返回 1, a[1]（除非 a[1]=nil）；第二次迭代调用 iter(a,1) 返回 2, a[2]……直到第一个 nil 元素。
【正文】
关于泛型for循环，有两个需要注意的问题。
如果仔细思考，可以发现这两个问题都已经被我标注出来了，就是上方的粗体斜体字。
这两个问题是：
1.到底怎么样才会让for退出？
2.什么叫“每次调用该函数就返回集合的下一个元素”
这篇文章说第一个，用户到底要做什么才能退出泛型for循环。
先看下列代码
> a = {} --成功，不报错
> a['s'] = 12 --成功，不报错
> a[2] = 3 --成功，不报错
> a[a] = 2 --成功，不报错
> a[foo] = 56 -- table index is nil
于是我们得到第一个结论：lua并不支持用户将nil类型的值作为key使用
再看下列代码
> a = {1,2,3,4,nil,'tyt','rrp',nil,20,30,'xss'}
> for k,v in ipairs(a) do
>> print(k,v)
>> end
1 1
2 2
3 3
4 4
> for k,v in pairs(a) do
>> print(k,v)
>> end
1 1
2 2
3 3
4 4
6 tyt
7 rrp
9 20
10 30
11 xss
在这里，lua直接跳过为value为nil的键值对，但并没有报错；不仅没报错，甚至还为这些值为nil的键值对计算了key，导致key出现断层。
于是我们可以得到第二个结论：lua支持nil作为表的value
所以可见，更有可能是当key为nil时退出for循环，而不是当value为nil时退出（即所谓的“a[1]=nil”“直到第一个nil元素”）
下面来证实一下这个说法。
a = {1,2,3,4,5,6,7}
function iter (a, i)
i = i + 1
local v = a[i]
if v then
return i, nil
end
end
function ipairs (a)
return iter, a, 0
end
for k,v in ipairs(a) do
print(k,v)
end
输出
1 nil
2 nil
3 nil
4 nil
5 nil
6 nil
7 nil
对比对照
a = {1,2,3,4,5,6,7}
function iter (a, i)
i = i + 1
local v = a[i]
if v then
return nil,v
end
end
function ipairs (a)
return iter, a, 0
end
for k,v in ipairs(a) do
print(k,v)
end
输出
--无
综上，当迭代器给value返回了一个nil值时，for循环继续运行；当迭代器给key返回了一个nil值时，for循环退出
那么正常的ipairs是怎么工作的呢？
for循环写好后，lua会优先计算右边表达式的值：ipairs返回给for三个信息，iter函数、要被迭代的对象（状态常量）、下标（控制变量）。之后，for开始执行iter函数。
iter函数执行中，当lua处理到分支语句时
1.如果v为nil则条件判断为false，那么就不执行return语句，因此i的值就未被返回到k上去，结果k的值被设为nil，所以跳出循环。（有两种不同的操作可以使k为nil，大概的原理见延伸2）
2.如果v不是nil，也就是说v有具体的值，那么执行return语句。因为i一定是一个number类变量，所以for会一直执行、循环，直到v为nil时
延伸：
1.要想for运行，那么迭代器返回的第一个元素就至关重要——第二个、第三个、第n个元素都是可有可无的，但第一个元素必定不能为nil
2.在for循环体内更改k、v的值并不能造成for循环退出。根据lua手册的解释
“环变量 var_i 对于循环来说是一个局部变量； 你不可以在 for 循环结束后继续使用。 如果你需要保留这些值，那么就在循环跳出或结束前赋值到别的变量里去”
我认为每进行一次循环，都相当于做了一次“local k,v = iter(状态常量，控制变量)”的操作。在这个操作中，lua已经做完了对于k是否为nil值这个问题的检测操作——这是lua自动进行的，是和for关键字绑定在一起的，所以在循环体内更改k、v并没有什么影响。
也因此，当iter没有返回值时，k会变成nil，具体的原理见延伸4.
3.如何证明每次循环都会新建同名本地变量
原理：在同一作用域中，使用local关键字定义同名变量时，lua不会将原有变量设为nil——也就是说原有变量会保留其原来的值，不会变成新定义时赋予的值，也不会变为nil——但此时该原有变量已无法手动访问，除非闭包。而如果该原有变量没有外部引用来为它形成闭包的话，它会被回收。
collectgarbage("stop")
for k,v in ipairs({1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9}) do
k = {1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,91,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8}
v = {1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,91,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8}
print(collectgarbage("count"))
end
for k,v in ipairs({1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9}) do
k = {1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,91,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8}
v = {1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,91,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8}
print(collectgarbage("count"))
end
for k,v in ipairs({1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9}) do
k = {1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,91,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8}
v = {1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,91,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8}
print(collectgarbage("count"))
end
for k,v in ipairs({1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9}) do
k = {1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,91,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8}
v = {1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,91,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8}
print(collectgarbage("count"))
end
for k,v in ipairs({1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9}) do
k = {1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,91,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8}
v = {1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,91,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8}
print(collectgarbage("count"))
end
for k,v in ipairs({1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9}) do
k = {1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,91,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8}
v = {1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,91,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8}
print(collectgarbage("count"))
end
for k,v in ipairs({1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9}) do
k = {1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,91,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8}
v = {1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,91,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8}
print(collectgarbage("count"))
end
for k,v in ipairs({1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9}) do
k = {1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,91,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8}
v = {1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,91,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8}
print(collectgarbage("count"))
end
collectgarbage("collect")
for k,v in ipairs({1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9}) do
k=nil
v=nil
print(collectgarbage("count"))
end
for k,v in ipairs({1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9}) do
k=nil
v=nil
print(collectgarbage("count"))
end
for k,v in ipairs({1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9}) do
k=nil
v=nil
print(collectgarbage("count"))
end
for k,v in ipairs({1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9}) do
k=nil
v=nil
print(collectgarbage("count"))
end
for k,v in ipairs({1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9}) do
k=nil
v=nil
print(collectgarbage("count"))
end
for k,v in ipairs({1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9}) do
k=nil
v=nil
print(collectgarbage("count"))
end
for k,v in ipairs({1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9}) do
k=nil
v=nil
print(collectgarbage("count"))
end
for k,v in ipairs({1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9}) do
k=nil
v=nil
print(collectgarbage("count"))
end
--[[
>lua -e "io.stdout:setvbuf 'no'" "定义全局变量和local定义变量肯定不是幂等的.lua"
41.6533203125
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62.7294921875
65.0712890625
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>Exit code: 0
]]--
4.为什么没有返回值时k会是nil？换句话说，要怎么样才能让上个循环中变量的信息不会保留到下一个循环中去？
for循环对于其局部变量k、v的定义可能等价于“local k,v = iter(状态常量，控制变量)”，也可能等价于“local k;local v;k,v = iter(状态常量，控制变量)”。这两种写法最终得到的结果是一样的，但其中有一些值得关注的细节。
local x会强制创造一个叫做x的nil值。
第一种写法直接让k、v等于iter的返回值，第二种写法则是先为它们赋值为nil再分配返回值。
（当然，这两种写法也有其他的区别。lua在发现一个函数类型的变量时，会先找局部变量再找全局变量。这种写法上的差异会造成递归bug。）
而对于“接受返回值”这个行为，看下列代码
> function f()
>> end
> x = f()
> print(x)
nil
> print(f())
--空
> y = 12
> y = f()
> print(y)
nil
>
> function f()
>> return nil
>> end
> print(f())
nil
>
看起来这是一个奇怪的bug，print可以打印x（nil值），却无法打印f()（不返回任何值），但我们也可以这样理解：当且仅当语句是一个赋值表达式，且表达式的右边还是一个没有返回值的函数（或者返回值为空，比如光写一个return，没有任何参数）时，那么lua会自动为表达式左边的变量赋上nil值。
菜鸟教程：若变量个数 > 值的个数 ，则按变量个数补足nil
不过看下列代码：
> function a()
>> end
> function f(i)
>> i = i + 1
>> end
> function t(...)
>> local args = {...}
>> print(args[1],args[2])
>> end
> t(nil,nil)
nil nil
> t(a(),a())
nil nil
> f(a())
stdin:2: attempt to perform arithmetic on local 'i' (a nil value)
stack traceback:
stdin:2: in function 'f'
stdin:1: in main chunk
[C]: ?
>
其实“赋值表达式”这个说法不准确，应该说是“发生赋值操作的语句”。
【附录】
> a = {false,false}
> for k,v in ipairs(a) do
>> print(k,v)
>> end
1 false
2 false
事实证明，只有为nil的value不会输出，为false的value仍然可以输出。
所以菜鸟教程上提供的这套迭代器实现原理是不准确的
