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])
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])