I tried to call function tan of math.h this way (directly copy the declaration) and it works:
local ffi = require("ffi")
ffi.cdef[[
double tan(double x);
]]
print(ffi.C.tan(45))
But when I tried to call the function localtime of time.h the same way:
local ffi = require("ffi")
ffi.cdef[[
struct tm *localtime(const time_t *tp);
]]
print(ffi.C.localtime(1234544))
And get error:
lua: C:\Users\xiang\Desktop\bm.lua:4: declaration specifier expected near 'time_t'
stack traceback:
[C]: in function 'cdef'
C:\Users\xiang\Desktop\bm.lua:4: in main chunk
[C]: at 0x00401f00
[Finished in 0.1s with exit code 1]
I've checked the official manual this and this but still confused.
Every function you would like to call from FFI, it needs to be defined before. If not LuaJIT does not how to parse a FFI function call, how to do data-type conversion from Lua to C (and viceversa), etc.
Keeping that in my mind, to make your code work you would need to define time_t and struct tm. time_t is generally defined as a signed integer. You can find the definition of struct tm in localtime docs (man localtime).
ffi.cdef[[
struct tm {
int tm_sec; /* Seconds (0-60) */
int tm_min; /* Minutes (0-59) */
int tm_hour; /* Hours (0-23) */
int tm_mday; /* Day of the month (1-31) */
int tm_mon; /* Month (0-11) */
int tm_year; /* Year - 1900 */
int tm_wday; /* Day of the week (0-6, Sunday = 0) */
int tm_yday; /* Day in the year (0-365, 1 Jan = 0) */
int tm_isdst; /* Daylight saving time */
};
struct tm *localtime(const int32_t *tp);
]]
In addition, function localtime expects a pointer value, not a constant integer. So it would be necessary to pass a c-data pointer storing an integer to localtime. There's a sort of LuaJIT idiom for that.
local time = ffi.new("int32_t[1]")
time[0] = 1234544
local tm = C.localtime(time)
Since arrays and pointers in C, although not the exactly same, are interchangeable in most cases.
Lastly, you cannot print a struct tm directly. Should store it into a variable and print out the fields you're interested.
print(tm.tm_sec)
You cannot use time_t as it isn't native C type. Replace it with proper native type or use a corresponding struct typedef. Then it should work.
Related
I need some help converting a luajit pointer to a string and back.
First I define the ctype:
ffi.cdef[[
typedef struct {
unsigned char Bytes[16];
} EncryptionKeys[100000000];
void* malloc(size_t);
void free(void*);
]]
Then use malloc to allocate some memory and then create the 'EncryptionKeys' variable.
local EncryptionKeyMemoryAddress = ffi.C.malloc(ffi.sizeof("EncryptionKeys"))
local EncryptionKeys = ffi.cast("EncryptionKeys(&)", EncryptionKeyMemoryAddress)
I first convert the variable into a lua string using:
ffi.string(EncryptionKeyMemoryAddress)
But I can't figure out how to convert it back!
Can someone please help me?
FYI: I am passing the 'EncryptionKeyMemoryAddress' variable to one of the function parameters for a lua lane (https://lualanes.github.io/lanes/).
Edit:
Here is the section of code that I am working on:
This is for the client managers module of my server that manages a list of lua states that all have access to any clients connected to the server. They all use a shared section of memory that I want them to have access to using a pointer.
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)
You can convert Lua string to your structure pointer (and later use it as an array):
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])
UPDATE:
Let's test how it works for an array containing three keys:
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])
It prints 11, 22, 33
UPDATE 2:
Pass the address of buffer instead of the content of buffer
In the main thread
-- 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
Inside the lane
-- 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])
I would like to have a C function return a string table array (e.g. {"a", "b", "c"}) to a Lua script via LuaJIT.
Which is the best way to do it?
I thought of returning a single concatenated string with some separator (e.g. "a|b|c") then splitting it in Lua, but I was wondering if there is a better way.
EDIT: I'm using LuaJIT FFI to call C functions.
I think the easiest way to accomplish this would be to have the C code return a struct containing an array of strings and a length to Lua and write a little Lua to reify it to your desired data structure.
In C:
typedef struct {
char *strings[];
size_t len;
} string_array;
string_array my_func(...) {
/* do what you are going to do here */
size_t nstrs = n; /* however many strings you are returning */
char** str_array = malloc(sizeof(char*)*nstrs);
/* put all your strings into the array here */
return {str_array, nstrs};
}
In Lua:
-- load my_func and string_array declarations
local str_array_C = C.ffi.my_func(...)
local str_array_lua = {}
for i = 0, str_array_C.len-1 do
str_array_lua[i+1] = ffi.string(str_array_C.strings[i])
end
-- str_array_lua now holds your list of strings
It always return a String which is (at least I guess) the table identifier someone can help in anyway?
Thats my function:
function listFiles(dir)
local ffi = require("ffi")
ffi.cdef[[char ** PHYSFS_enumerateFiles ( const char * dir );]]
local liblove = ffi.os == "Windows" and ffi.load("love") or ffi.C
local tb={}
tb=liblove.PHYSFS_enumerateFiles(dir)
return tb
end
It should return me a String with filecontents of the "Dir" I pass to it, but it doesnt. Can't figure out why.
You should read the reference properly. The enumeration function returns a pointer to string pointers, after the last string follows a NULL pointer. Conversion from char* to Lua string can be done with ffi.string.
I have a trace file that each transaction time represented in Windows filetime format. These time numbers are something like this:
128166372003061629
128166372016382155
128166372026382245
Would you please let me know if there are any C/C++ library in Unix/Linux to extract actual time (specially second) from these numbers ? May I write my own extraction function ?
it's quite simple: the windows epoch starts 1601-01-01T00:00:00Z. It's 11644473600 seconds before the UNIX/Linux epoch (1970-01-01T00:00:00Z). The Windows ticks are in 100 nanoseconds. Thus, a function to get seconds from the UNIX epoch will be as follows:
#define WINDOWS_TICK 10000000
#define SEC_TO_UNIX_EPOCH 11644473600LL
unsigned WindowsTickToUnixSeconds(long long windowsTicks)
{
return (unsigned)(windowsTicks / WINDOWS_TICK - SEC_TO_UNIX_EPOCH);
}
FILETIME type is is the number 100 ns increments since January 1 1601.
To convert this into a unix time_t you can use the following.
#define TICKS_PER_SECOND 10000000
#define EPOCH_DIFFERENCE 11644473600LL
time_t convertWindowsTimeToUnixTime(long long int input){
long long int temp;
temp = input / TICKS_PER_SECOND; //convert from 100ns intervals to seconds;
temp = temp - EPOCH_DIFFERENCE; //subtract number of seconds between epochs
return (time_t) temp;
}
you may then use the ctime functions to manipulate it.
(I discovered I can't enter readable code in a comment, so...)
Note that Windows can represent times outside the range of POSIX epoch times, and thus a conversion routine should return an "out-of-range" indication as appropriate. The simplest method is:
... (as above)
long long secs;
time_t t;
secs = (windowsTicks / WINDOWS_TICK - SEC_TO_UNIX_EPOCH);
t = (time_t) secs;
if (secs != (long long) t) // checks for truncation/overflow/underflow
return (time_t) -1; // value not representable as a POSIX time
return t;
New answer for old question.
Using C++11's <chrono> plus this free, open-source library:
https://github.com/HowardHinnant/date
One can very easily convert these timestamps to std::chrono::system_clock::time_point, and also convert these timestamps to human-readable format in the Gregorian calendar:
#include "date.h"
#include <iostream>
std::chrono::system_clock::time_point
from_windows_filetime(long long t)
{
using namespace std::chrono;
using namespace date;
using wfs = duration<long long, std::ratio<1, 10'000'000>>;
return system_clock::time_point{floor<system_clock::duration>(wfs{t} -
(sys_days{1970_y/jan/1} - sys_days{1601_y/jan/1}))};
}
int
main()
{
using namespace date;
std::cout << from_windows_filetime(128166372003061629) << '\n';
std::cout << from_windows_filetime(128166372016382155) << '\n';
std::cout << from_windows_filetime(128166372026382245) << '\n';
}
For me this outputs:
2007-02-22 17:00:00.306162
2007-02-22 17:00:01.638215
2007-02-22 17:00:02.638224
On Windows, you can actually skip the floor, and get that last decimal digit of precision:
return system_clock::time_point{wfs{t} -
(sys_days{1970_y/jan/1} - sys_days{1601_y/jan/1})};
2007-02-22 17:00:00.3061629
2007-02-22 17:00:01.6382155
2007-02-22 17:00:02.6382245
With optimizations on, the sub-expression (sys_days{1970_y/jan/1} - sys_days{1601_y/jan/1}) will translate at compile time to days{134774} which will further compile-time-convert to whatever units the full-expression requires (seconds, 100-nanoseconds, whatever). Bottom line: This is both very readable and very efficient.
The solution that divides and adds will not work correctly with daylight savings.
Here is a snippet that works, but it is for windows.
time_t FileTime_to_POSIX(FILETIME ft)
{
FILETIME localFileTime;
FileTimeToLocalFileTime(&ft,&localFileTime);
SYSTEMTIME sysTime;
FileTimeToSystemTime(&localFileTime,&sysTime);
struct tm tmtime = {0};
tmtime.tm_year = sysTime.wYear - 1900;
tmtime.tm_mon = sysTime.wMonth - 1;
tmtime.tm_mday = sysTime.wDay;
tmtime.tm_hour = sysTime.wHour;
tmtime.tm_min = sysTime.wMinute;
tmtime.tm_sec = sysTime.wSecond;
tmtime.tm_wday = 0;
tmtime.tm_yday = 0;
tmtime.tm_isdst = -1;
time_t ret = mktime(&tmtime);
return ret;
}
Assuming you are asking about the FILETIME Structure, then FileTimeToSystemTime does what you want, you can get the seconds from the SYSTEMTIME structure it produces.
Here's essentially the same solution except this one encodes negative numbers from Ldap properly and lops off the last 7 digits before conversion.
public static int LdapValueAsUnixTimestamp(SearchResult searchResult, string fieldName)
{
var strValue = LdapValue(searchResult, fieldName);
if (strValue == "0") return 0;
if (strValue == "9223372036854775807") return -1;
return (int)(long.Parse(strValue.Substring(0, strValue.Length - 7)) - 11644473600);
}
If somebody need convert it in MySQL
SELECT timestamp,
FROM_UNIXTIME(ROUND((((timestamp) / CAST(10000000 AS UNSIGNED INTEGER)))
- CAST(11644473600 AS UNSIGNED INTEGER),0))
AS Converted FROM events LIMIT 100
Also here's a pure C#ian way to do it.
(Int32)(DateTime.FromFileTimeUtc(129477880901875000).Subtract(new DateTime(1970, 1, 1))).TotalSeconds;
Here's the result of both methods in my immediate window:
(Int32)(DateTime.FromFileTimeUtc(long.Parse(strValue)).Subtract(new DateTime(1970, 1, 1))).TotalSeconds;
1303314490
(int)(long.Parse(strValue.Substring(0, strValue.Length - 7)) - 11644473600)
1303314490
DateTime.FromFileTimeUtc(long.Parse(strValue))
{2011-04-20 3:48:10 PM}
Date: {2011-04-20 12:00:00 AM}
Day: 20
DayOfWeek: Wednesday
DayOfYear: 110
Hour: 15
InternalKind: 4611686018427387904
InternalTicks: 634389112901875000
Kind: Utc
Millisecond: 187
Minute: 48
Month: 4
Second: 10
Ticks: 634389112901875000
TimeOfDay: {System.TimeSpan}
Year: 2011
dateData: 5246075131329262904
In my previous question I was looking for a way of evaulating complex mathematical expressions in C, most of the suggestions required implementing some type of parser.
However one answer, suggested using Lua for evaluating the expression. I am interested in this approach but I don't know anything about Lua.
Can some one with experience in Lua shed some light?
Specifically what I'd like to know is
Which API if any does Lua provide that can evaluate mathematical expressions passed in as a string? If there is no API to do such a thing, may be some one can shed some light on the linked answer as it seemed like a good approach :)
Thanks
The type of expression I'd like to evaluate is given some user input such as
y = x^2 + 1/x - cos(x)
evaluate y for a range of values of x
It is straightforward to set up a Lua interpreter instance, and pass it expressions to be evaluated, getting back a function to call that evaluates the expression. You can even let the user have variables...
Here's the sample code I cooked up and edited into my other answer. It is probably better placed on a question tagged Lua in any case, so I'm adding it here as well. I compiled this and tried it for a few cases, but it certainly should not be trusted in production code without some attention to error handling and so forth. All the usual caveats apply here.
I compiled and tested this on Windows using Lua 5.1.4 from Lua for Windows. On other platforms, you'll have to find Lua from your usual source, or from www.lua.org.
Update: This sample uses simple and direct techniques to hide the full power and complexity of the Lua API behind as simple as possible an interface. It is probably useful as-is, but could be improved in a number of ways.
I would encourage readers to look into the much more production-ready ae library by lhf for code that takes advantage of the API to avoid some of the quick and dirty string manipulation I've used. His library also promotes the math library into the global name space so that the user can say sin(x) or 2 * pi without having to say math.sin and so forth.
Public interface to LE
Here is the file le.h:
/* Public API for the LE library.
*/
int le_init();
int le_loadexpr(char *expr, char **pmsg);
double le_eval(int cookie, char **pmsg);
void le_unref(int cookie);
void le_setvar(char *name, double value);
double le_getvar(char *name);
Sample code using LE
Here is the file t-le.c, demonstrating a simple use of this library. It takes its single command-line argument, loads it as an expression, and evaluates it with the global variable x changing from 0.0 to 1.0 in 11 steps:
#include <stdio.h>
#include "le.h"
int main(int argc, char **argv)
{
int cookie;
int i;
char *msg = NULL;
if (!le_init()) {
printf("can't init LE\n");
return 1;
}
if (argc<2) {
printf("Usage: t-le \"expression\"\n");
return 1;
}
cookie = le_loadexpr(argv[1], &msg);
if (msg) {
printf("can't load: %s\n", msg);
free(msg);
return 1;
}
printf(" x %s\n"
"------ --------\n", argv[1]);
for (i=0; i<11; ++i) {
double x = i/10.;
double y;
le_setvar("x",x);
y = le_eval(cookie, &msg);
if (msg) {
printf("can't eval: %s\n", msg);
free(msg);
return 1;
}
printf("%6.2f %.3f\n", x,y);
}
}
Here is some output from t-le:
E:...>t-le "math.sin(math.pi * x)"
x math.sin(math.pi * x)
------ --------
0.00 0.000
0.10 0.309
0.20 0.588
0.30 0.809
0.40 0.951
0.50 1.000
0.60 0.951
0.70 0.809
0.80 0.588
0.90 0.309
1.00 0.000
E:...>
Implementation of LE
Here is le.c, implementing the Lua Expression evaluator:
#include <lua.h>
#include <lauxlib.h>
#include <stdlib.h>
#include <string.h>
static lua_State *L = NULL;
/* Initialize the LE library by creating a Lua state.
*
* The new Lua interpreter state has the "usual" standard libraries
* open.
*/
int le_init()
{
L = luaL_newstate();
if (L)
luaL_openlibs(L);
return !!L;
}
/* Load an expression, returning a cookie that can be used later to
* select this expression for evaluation by le_eval(). Note that
* le_unref() must eventually be called to free the expression.
*
* The cookie is a lua_ref() reference to a function that evaluates the
* expression when called. Any variables in the expression are assumed
* to refer to the global environment, which is _G in the interpreter.
* A refinement might be to isolate the function envioronment from the
* globals.
*
* The implementation rewrites the expr as "return "..expr so that the
* anonymous function actually produced by lua_load() looks like:
*
* function() return expr end
*
*
* If there is an error and the pmsg parameter is non-NULL, the char *
* it points to is filled with an error message. The message is
* allocated by strdup() so the caller is responsible for freeing the
* storage.
*
* Returns a valid cookie or the constant LUA_NOREF (-2).
*/
int le_loadexpr(char *expr, char **pmsg)
{
int err;
char *buf;
if (!L) {
if (pmsg)
*pmsg = strdup("LE library not initialized");
return LUA_NOREF;
}
buf = malloc(strlen(expr)+8);
if (!buf) {
if (pmsg)
*pmsg = strdup("Insufficient memory");
return LUA_NOREF;
}
strcpy(buf, "return ");
strcat(buf, expr);
err = luaL_loadstring(L,buf);
free(buf);
if (err) {
if (pmsg)
*pmsg = strdup(lua_tostring(L,-1));
lua_pop(L,1);
return LUA_NOREF;
}
if (pmsg)
*pmsg = NULL;
return luaL_ref(L, LUA_REGISTRYINDEX);
}
/* Evaluate the loaded expression.
*
* If there is an error and the pmsg parameter is non-NULL, the char *
* it points to is filled with an error message. The message is
* allocated by strdup() so the caller is responsible for freeing the
* storage.
*
* Returns the result or 0 on error.
*/
double le_eval(int cookie, char **pmsg)
{
int err;
double ret;
if (!L) {
if (pmsg)
*pmsg = strdup("LE library not initialized");
return 0;
}
lua_rawgeti(L, LUA_REGISTRYINDEX, cookie);
err = lua_pcall(L,0,1,0);
if (err) {
if (pmsg)
*pmsg = strdup(lua_tostring(L,-1));
lua_pop(L,1);
return 0;
}
if (pmsg)
*pmsg = NULL;
ret = (double)lua_tonumber(L,-1);
lua_pop(L,1);
return ret;
}
/* Free the loaded expression.
*/
void le_unref(int cookie)
{
if (!L)
return;
luaL_unref(L, LUA_REGISTRYINDEX, cookie);
}
/* Set a variable for use in an expression.
*/
void le_setvar(char *name, double value)
{
if (!L)
return;
lua_pushnumber(L,value);
lua_setglobal(L,name);
}
/* Retrieve the current value of a variable.
*/
double le_getvar(char *name)
{
double ret;
if (!L)
return 0;
lua_getglobal(L,name);
ret = (double)lua_tonumber(L,-1);
lua_pop(L,1);
return ret;
}
Remarks
The above sample consists of 189 lines of code total, including a spattering of comments, blank lines, and the demonstration. Not bad for a quick function evaluator that knows how to evaluate reasonably arbitrary expressions of one variable, and has rich library of standard math functions at its beck and call.
You have a Turing-complete language underneath it all, and it would be an easy extension to allow the user to define complete functions as well as to evaluate simple expressions.
Since you're lazy, like most programmers, here's a link to a simple example that you can use to parse some arbitrary code using Lua. From there, it should be simple to create your expression parser.
This is for Lua users that are looking for a Lua equivalent of "eval".
The magic word used to be loadstring but it is now, since Lua 5.2, an upgraded version of load.
i=0
f = load("i = i + 1") -- f is a function
f() ; print(i) -- will produce 1
f() ; print(i) -- will produce 2
Another example, that delivers a value :
f=load('return 2+3')
print(f()) -- print 5
As a quick-and-dirty way to do, you can consider the following equivalent of eval(s), where s is a string to evaluate :
load(s)()
As always, eval mechanisms should be avoided when possible since they are expensive and produce a code difficult to read.
I personally use this mechanism with LuaTex/LuaLatex to make math operations in Latex.
The Lua documentation contains a section titled The Application Programming Interface which describes how to call Lua from your C program. The documentation for Lua is very good and you may even be able to find an example of what you want to do in there.
It's a big world in there, so whether you choose your own parsing solution or an embeddable interpreter like Lua, you're going to have some work to do!
function calc(operation)
return load("return " .. operation)()
end