在 Lua 中，将未定义的标识符分配给未声明的变量是什么意思？

发布于 2024-09-27 22:18:42 字数 875 浏览 8 评论 0原文

我正在仔细阅读 Lua 代码文件，文件的最顶部包含：

    1  | TradeSkillFrameReset = TradeSkillFrame_LoadUI;
    2  | 
    3  | TradeSkillFrame_LoadUI = function()
    4  |    TradeSkillFrameReset();
            ...
    112|    TradeSkillFrame_LoadUI = TradeSkillFrameReset;
            ...
    114| end;

他们正在执行分配的第一行：

TradeSkillFrameReset = TradeSkillFrame_LoadUI;

此时在代码文件中，他们正在分配一个未声明的标识符 (TradeSkillFrame_LoadUI)未声明的变量 (TradeSkillFrameReset)。 Lua 允许这样做吗？

变量声明是隐式的吗？
所有未声明的标识符都被假定为前向声明吗？

更令人困惑的是，第一个 TradeSkillFrame_LoadUI 似乎是调用自身：

TradeSkillFrame_LoadUI = function()
   TradeSkillFrameReset();

这怎么不是一个无限循环？

编辑：添加了对 TradeSkillFrameReset 和 TradeSkillFrame_LoadUI 的剩余引用。

原文

I was perusing an Lua code file, and the very top of the file contains:

    1  | TradeSkillFrameReset = TradeSkillFrame_LoadUI;
    2  | 
    3  | TradeSkillFrame_LoadUI = function()
    4  |    TradeSkillFrameReset();
            ...
    112|    TradeSkillFrame_LoadUI = TradeSkillFrameReset;
            ...
    114| end;

The very first line they are doing an assignment:

TradeSkillFrameReset = TradeSkillFrame_LoadUI;

At this point in the code file they are assigning an undeclaraed identifier (TradeSkillFrame_LoadUI) to an undeclared variable (TradeSkillFrameReset). Is this allowed in Lua?

is variable declaration implicit?
are all undeclared identifiers assumed to be forward declarations?

The more confusing thing is that the first TradeSkillFrame_LoadUI seems to do is call itself:

TradeSkillFrame_LoadUI = function()
   TradeSkillFrameReset();

How is this not an infinite loop?

Edit: Added remaining references to TradeSkillFrameReset and TradeSkillFrame_LoadUI.

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梦巷 2024-10-04 22:18:42

如果 TradeSkillFrame_LoadUI 是全局函数，则第一条语句会在 TradeSkillFrameReset 中捕获此函数。

然后，对 TradeSkillFrame_LoadUI 的分配会将全局函数替换为新函数，该函数首先通过 TradeSkillFrameReset 引用调用原始函数。

这种模式被称为“函数挂钩”，这里是一些关于通用技术和一些魔兽世界的更多信息Lua 环境特定详细信息（根据函数名称，该脚本可能来自何处）

一些示例代码，以明确这一点：

function SomeGlobalFunction()
    print("I'm the original global function")
end

OrigFunction = SomeGlobalFunction

SomeGlobalFunction = function()
    OrigFunction()
    print("And here's the modfied one...")
end

--...
SomeGlobalFunction()

这将打印以下输出：

我是原来的全局函数
这是修改后的...

顺便说一下，Lua 的名字是并非全部大写。

If TradeSkillFrame_LoadUI is a global function, the first statement captures this function in TradeSkillFrameReset.

The assigment to TradeSkillFrame_LoadUI then replaces the global function with a new one, that at first calls the original function through the TradeSkillFrameReset reference.

This pattern is called "function hooking", here is some more information about the general technique and some World of Warcraft Lua environment specific details (where this script may come from, according to the function's name)

Some example code, to make this clear:

function SomeGlobalFunction()
    print("I'm the original global function")
end

OrigFunction = SomeGlobalFunction

SomeGlobalFunction = function()
    OrigFunction()
    print("And here's the modfied one...")
end

--...
SomeGlobalFunction()

this will print the following output:

I'm the original global function
And here's the modfied one...

As a side note, Lua's name is not all capitalized.

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神爱温柔 2024-10-04 22:18:42

在第一行，由于 TradeSkillFrame_LoadUI 是一个未初始化的变量，因此第一行：

TradeSkillFrameReset = TradeSkillFrame_LoadUI;

与以下代码相同：

TradeSkillFrameReset = nil

下面的函数不会进入无限循环，因为 TradeSkillFrameReset 实际上不是 '指向那里的任何东西。

我的猜测是，稍后在代码中，它会被正确初始化。就像这样：

TradeSkillFrameReset = TradeSkillFrame_LoadUI;

TradeSkillFrame_LoadUI = function()  -- 1. function declaration
   TradeSkillFrameReset();

   [...snip...]
end;

[... snip ...]
TradeSkillFrameReset = somethingElse -- 2. initialized to something else
TradeSkillFrame_LoadUI()             -- 3. function invocation

技巧是你可以在 1 上定义一个函数，并在 3 上使用变量的值，同时在 2 上更改它。

如果这还不够清楚，请告诉我。

On the first line, since TradeSkillFrame_LoadUI is a non-initialized variable, then this first line:

TradeSkillFrameReset = TradeSkillFrame_LoadUI;

Is the same as making:

TradeSkillFrameReset = nil

The function below that doesn't enter an infinite loop because TradeSkillFrameReset isn't actually 'pointing to anything' there.

My guess is that later on in the code, it gets properly initialized. Like so:

TradeSkillFrameReset = TradeSkillFrame_LoadUI;

TradeSkillFrame_LoadUI = function()  -- 1. function declaration
   TradeSkillFrameReset();

   [...snip...]
end;

[... snip ...]
TradeSkillFrameReset = somethingElse -- 2. initialized to something else
TradeSkillFrame_LoadUI()             -- 3. function invocation

The trick is that you can define a function on 1, and use the value of a variable on 3, while changing it on 2.

Let me know if this is not clear enough.

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任谁 2024-10-04 22:18:42

除非您正在使用的环境中全局变量的行为已被修改，否则该语句绝对不会执行任何操作。

local TradeSkillFrameReset = TradeSkillFrame_LoadUI

会产生效果，创建一个局部变量“槽”并使代码超过该点使用局部变量槽进行 TradeSkillFrameReset，而不是对其进行全局变量查找。然而，正如您发布的代码中一样，它只是将全局变量分配给 nil，这具有擦除它的效果，换句话说，如果它已经是 nil 则没有效果。

可能存在的原因：

1) 完成全局变量赋值后，代码运行的环境会执行一些特殊操作，而不是简单赋值的默认行为。但我认为这不太可能，就好像有任何特殊行为一样，查找 nil TradeSkillFrame_LoadUI 可能会导致错误。

2）最可能的原因只是为了可读性。它是为了让您知道 TradeSkillFrameReset将被正确分配到稍后埋藏在代码中的某个地方，您不会轻易注意到它。

Unless the behavior of global variables has been modified in the environment that you're using has been modified, the statement does absolutely nothing.

local TradeSkillFrameReset = TradeSkillFrame_LoadUI

Would have an effect, creating a local variable "slot" and making code past that point use the local variable slot for TradeSkillFrameReset rather than doing a global variable lookup for it. However, as it is in the code you posted, it simply assigns the global variable to nil, which has the effect of erasing it, in other words, no effect if it was already nil.

Reasons it could be there:

1) The environment the code is running in does something special when a global variable assignment is done, rather than the default behavior of a simple assignment. I don't think this is likely though, as if there was any special behavior the lookup of the nil TradeSkillFrame_LoadUI would probably cause an error.

2) The most probable reason is simply for readability. It's to let you know that TradeSkillFrameReset will be properly assigned to later buried somewhere in the code where you won't notice it as readily.

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