4

我在这里看到了如何使用 Lua (5.1) 中的 debug.getlocal 函数在表中插入局部变量。

function locals()
   local variables = {}
   local idx = 1
   while true do
     local ln, lv = debug.getlocal(2, idx)
     if ln ~= nil then
       variables[ln] = lv
     else
       break
     end
     idx = 1 + idx
   end
   return variables
 end

但是，当我尝试返回创建的表并访问它的条目时，它不起作用。

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

print(test1().v) -- nil

经过一些跟踪和错误后，我注意到在返回之前将表绑定到变量，或者简单地添加括号可以修复行为：

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

print(test2().v) -- 'I am a local!'

这让我很困惑。为什么这两个程序有任何不同？我不明白什么？locals() 处于尾调用位置这一事实有什么不同吗？

4

1 回答 1

4

我想让你感到困惑的是lua的正确尾调用特性。

为了理解这一点，我们修改了您的locals函数，使其接受一个参数作为调用中使用的级别堆栈debug.getlocal。（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 = 3.
key = level, val = 1.

What test3 returns: 
key = variables, val = table: 0x7fa14bc088b0.
key = idx, val = 3.
key = level, 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 = 2.
key = v, val = I am a local!.

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

What test2 returns: 
key = (for step), val = 1.
key = (for limit), val = 3.
key = (for index), val = 2.
key = level, val = 3.
key = printTable, val = function: 0x7fa14bc08360.
key = (*temporary), val = function: 0x7fa14bc08360.
key = num, val = 2.

What test3 returns: 
key = (for step), val = 1.
key = (for limit), val = 3.
key = (for index), val = 3.
key = level, val = 3.
key = printTable, val = function: 0x7fa14bc08360.
key = (*temporary), val = function: 0x7fa14bc08360.
key = num, val = 3.

何时level为 1，locals可以很好地提供自己的局部变量。但是当level为 2 时，test1返回外部范围的变量，而test2并test3给出您期望的结果。对于堆栈级别 3test2并test3返回类似于test1堆栈级别 2 的内容。所以它似乎test1 跳过了堆栈级别，我能想到的唯一解释是正确的尾调用。

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，我认为这是以某种内联方式进行调用。如果我是对的，这解释了test1return 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。

于 2016-08-06T05:25:14.183 回答