与Lua中的debug.getlocal混淆

为了理解这一点,我们修改了locals函数,使其接受一个参数作为调用debug.getlocal时使用的级别堆栈. (我正在使用Lua 5.3.3)

-- Get local variables with stack level 'level'.
function locals(level)
    local variables = {}
    local idx = 1
    while true do
        local ln,lv = debug.getlocal(level,idx)
        if ln ~= nil then
            variables[ln] = lv
        else
            break
        end
        idx = 1 + idx
    end
    return variables
end

然后我们修改你的测试函数,添加相同的参数,并添加一个test3函数以供参考.

function test1(level)
    local v = 'I am a local!'
    return locals(level)
end

function test2(level)
    local v = 'I am a local!'
    return (locals(level))
end

function test3(level)
    local v = 'I am a local!'
    local a = locals(level)
    return a
end

最后,我们添加一些代码来运行测试.

local function printTable(t)
    -- print(t)
    for k,v in pairs(t) do
        print(string.format("key = %s,val = %s.",k,v))
    end
end

for level = 1,3 do
    print("==== Stack level: " .. tostring(level))
    for num = 1,3 do
        print(string.format("What test%d returns: ",num))
        printTable(_G[(string.format("test%d",num))](level))
        print("")
    end
end

上面的代码运行具有不同堆栈级别的测试函数,并打印返回的键值对.我的结果如下：

==== Stack level: 1
What test1 returns: 
key = variables,val = table: 0x7fa14bc081e0.
key = idx,val = 3.
key = level,val = 1.

What test2 returns: 
key = variables,val = table: 0x7fa14bc08220.
key = idx,val = 1.

What test3 returns: 
key = variables,val = table: 0x7fa14bc088b0.
key = idx,val = 1.

==== Stack level: 2
What test1 returns: 
key = (for step),val = 1.
key = (for limit),val = 3.
key = (for index),val = 1.
key = level,val = 2.
key = printTable,val = function: 0x7fa14bc08360.
key = (*temporary),val = function: 0x7fa14bc08360.
key = num,val = 1.

What test2 returns: 
key = level,val = 2.
key = v,val = I am a local!.

What test3 returns: 
key = level,val = I am a local!.

==== Stack level: 3
What test1 returns: 
key = (*temporary),val = function: 0x109f5a070.

What test2 returns: 
key = (for step),val = 2.
key = level,val = 3.
key = printTable,val = 2.

What test3 returns: 
key = (for step),val = 3.

当level为1时,locals可以很好地给出它自己的局部变量.但是当level为2时,test1返回外部作用域的变量,而test2和test3给出你期望的结果.对于堆栈级别3,test2和test3在堆栈级别2返回类似test1的内容.因此,似乎test1跳过堆栈级别,我能想到的唯一解释是正确的尾部调用.

根据PIL(我在开头提供的链接),正确的尾调用永远不会导致堆栈溢出,我采用一些内联方式进行调用.如果我是对的,这解释了test1的return语句的跳过行为,因为这是一个正确的尾调用,并且是3个测试函数中唯一的一个.