如何将 luajit 指针转换为字符串并返回?

How to convert a luajit pointer to a string and back?

我需要一些帮助将 luajit 指针转换为字符串并返回。

首先我定义了ctype:

ffi.cdef[[
    typedef struct {
        unsigned char Bytes[16];
    } EncryptionKeys[100000000];

void* malloc(size_t);                   
void free(void*);
]]

然后使用 malloc 分配一些内存,然后创建 'EncryptionKeys' 变量。

local EncryptionKeyMemoryAddress = ffi.C.malloc(ffi.sizeof("EncryptionKeys"))

local EncryptionKeys = ffi.cast("EncryptionKeys(&)", EncryptionKeyMemoryAddress)

我首先使用以下方法将变量转换为 lua 字符串:

ffi.string(EncryptionKeyMemoryAddress)

但我不知道如何将它转换回来! 有人可以帮我吗?

仅供参考:我将 'EncryptionKeyMemoryAddress' 变量传递给 lua 车道 (https://lualanes.github.io/lanes/) 的函数参数之一。

编辑: 这是我正在处理的代码部分: 这是我服务器的客户端管理器模块,该模块管理 lua 状态列表,所有状态都可以访问连接到服务器的任何客户端。他们都使用我希望他们可以使用指针访问的共享内存部分。



local ClientFFIString = [[
    
    typedef struct {
        unsigned char Bytes[16];
    } EncryptionKeys[100000000];

    void* malloc(size_t);                   
    void free(void*);
]]

ffi.cdef(Matchpools.FFIString)

local EncryptionKeyMemoryAddress = ffi.C.malloc(ffi.sizeof("EncryptionKeys"))

--------------------------------------------
function ClientManagers.CreateNewClientManager()

    local EncryptionKeys = ffi.cast("EncryptionKeys(&)", EncryptionKeyMemoryAddress)

    EncryptionKeys[0].Bytes[0] = 24

    print("___a", EncryptionKeys[0].Bytes[0])


    local NewIndex = #ClientManagers.List+1
    ClientManagers.List[NewIndex] = ClientManagerFunc(
        ClientFFIString, 
        ffi.string(EncryptionKeysMemoryAddress)
    )

end


--------------------------------------------
local ClientManagerFunc = Lanes.gen("*", function(ClientFFIString, EncryptionKeysMemoryAddress)

    ffi = require("ffi")

    ffi.cdef(ClientFFIString)


    local EncryptionKeys = ffi.cast("EncryptionKeys(&)", EncryptionKeyMemoryAddress)
    
    print("___a", EncryptionKeys[0].Bytes[0]) 
    -- I want this to be 24 just like it is in the function that created this lua state


    local ClientManagerRunning = true
    while ClientManagerRunning do

        --local dt = GetDt()

        --UpdateClientData(dt)

        --UpdateMatchmaking(dt)

    end

end)

您可以将 Lua 字符串转换为您的结构指针(稍后将其用作数组):

ffi.cdef"typedef struct {unsigned char Bytes[16];} Key;"
ptr=ffi.cast("Key *", your_string)
print("First Byte of the First Key in your Array:", ptr[0].Bytes[0])

更新:
让我们测试它如何处理包含三个键的数组:

local ffi = require'ffi'
ffi.cdef"typedef struct {unsigned char Bytes[16];} Key;"
local your_string = string.char(11):rep(16)..string.char(22):rep(16)..string.char(33):rep(16)
local ptr=ffi.cast("Key *", your_string)
print("First Byte of the First Key in your Array:", ptr[0].Bytes[0])
print("First Byte of the Second Key in your Array:", ptr[1].Bytes[0])
print("First Byte of the Third Key in your Array:", ptr[2].Bytes[0])

它打印出 11、22、33


更新 2:
传递缓冲区的地址而不是缓冲区的内容

在主线程中

-- allocate the buffer
local Keys = ffi.cast("EncryptionKeys&", ffi.C.malloc(ffi.sizeof("EncryptionKeys")))
-- write to the buffer
Keys[2].Bytes[5] = 42
-- create string containing 64-bit address
local string_to_send = tostring(ffi.cast("uint64_t", Keys))
-- Send string_to_send to a lane

车道内

-- receive the string
local received_string = .....
-- restore the buffer pointer
local Keys = ffi.cast("EncryptionKeys&", loadstring("return "..received_string)())
-- read the data from the buffer
print(Keys[2].Bytes[5])