The code is:
#include <iostream>
#include <opencv2/opencv.hpp>
int main (int argc, char** argv) {
auto path = "C:/Users/huhua/Pictures/11.jpg";
auto img = cv::imread(path);
if (img.empty()) {
std::cout << "is empty" << std::endl;
return 1;
}
cv::imshow("demo", img);
cv::waitKey(0);
return 0;
}
The 11.jpg exist. And if I use another 11.bmp. It works well.
After debug. The error is throw at libjpeg-trubo/src/jdatasr.c
fill_input_buffer(j_decompress_ptr cinfo)
{
my_src_ptr src = (my_src_ptr)cinfo->src;
size_t nbytes;
// error is throw at here
nbytes = JFREAD(src->infile, src->buffer, INPUT_BUF_SIZE);
// ...
}
Is my libjpeg issue??
How to fix this?
The 11.jpg image:
Update:
The OpenCV info
Update on 2021/10/19:
The reason is I set the cmake_toolchain_path after project(xxx). I should set the cmake_toolchain_path before project.
https://github.com/microsoft/vcpkg/discussions/20802
All examples and books I've seen so far recommends using waitKey(1) to force repaint OpenCV window. That looks weird and too hacky. Why wait for even 1ms when you don't have to?
Are there any alternatives? I tried cv::updateWindow but it seems to require OpenGL and therefore crashes. I'm using VC++ on Windows.
I looked in to source and as #Dan Masek said, there doesn't seem to be any other functions to process windows message. So I ended up writing my own little DoEvents() function for VC++. Below is the full source code that uses OpenCV to display video frame by frame while skipping desired number of frames.
#include <windows.h>
#include <iostream>
#include "opencv2/opencv.hpp"
using namespace cv;
using namespace std;
bool DoEvents();
int main(int argc, char *argv[])
{
VideoCapture cap(argv[1]);
if (!cap.isOpened())
return -1;
namedWindow("tree", CV_GUI_EXPANDED | CV_WINDOW_AUTOSIZE);
double frnb(cap.get(CV_CAP_PROP_FRAME_COUNT));
std::cout << "frame count = " << frnb << endl;
for (double fIdx = 0; fIdx < frnb; fIdx += 50) {
Mat frame;
cap.set(CV_CAP_PROP_POS_FRAMES, fIdx);
bool success = cap.read(frame);
if (!success) {
cout << "Cannot read frame " << endl;
break;
}
imshow("tree", frame);
if (!DoEvents())
return 0;
}
return 0;
}
bool DoEvents()
{
MSG msg;
BOOL result;
while (::PeekMessage(&msg, NULL, 0, 0, PM_NOREMOVE))
{
result = ::GetMessage(&msg, NULL, 0, 0);
if (result == 0) // WM_QUIT
{
::PostQuitMessage(msg.wParam);
return false;
}
else if (result == -1)
return true; //error occured
else
{
::TranslateMessage(&msg);
::DispatchMessage(&msg);
}
}
return true;
}
Is this the correct way to sync threads without mutex.
This code should be running for a long time
#include <boost/thread.hpp>
#include <boost/thread/mutex.hpp>
#include <boost/memory_order.hpp>
#include <atomic>
std::atomic<long> x =0;
std::atomic<long> y =0;
boost::mutex m1;
// Thread increments
void Thread_Func()
{
for(;;)
{
// boost::mutex::scoped_lock lx(m1);
++x;
++y;
}
}
// Checker Thread
void Thread_Func_X()
{
for(;;)
{
// boost::mutex::scoped_lock lx(m1);
if(y > x)
{
// should never hit until int overflows
std::cout << y << "\\" << x << std::endl;
break;
}
}
}
//Test Application
int main(int argc, char* argv[])
{
boost::thread_group threads;
threads.create_thread(Thread_Func);
threads.create_thread(Thread_Func_X);
threads.join_all();
return 0;
}
Without knowing exactly what you're trying to do, it is hard to say it is the "correct" way. That's valid code, it's a bit janky though.
There is no guarantee that the "Checker" thread will ever see the condition y > x. It's theoretically possible that it will never break. In practice, it will trigger at some point but x might not be LONG_MIN and y LONG_MAX. In other words, it's not guaranteed to trigger just as the overflow happens.
I just want to resume the func coroutine twice, yield if n==0, and return if n==1 , but it core dumps, what't wrong with it?
the "hello world" should always be left in LL's stack, I can't figure out what is wrong.
[liangdong#cq01-clientbe-code00.vm.baidu.com lua]$ ./main
func_top=1 top=hello world
first_top=1 top_string=hello world
Segmentation fault (core dumped)
[liangdong#cq01-clientbe-code00.vm.baidu.com lua]$ cat main.c
#include "lua.h"
#include "lualib.h"
#include "lauxlib.h"
int n = 0;
int func(lua_State *L) {
printf("func_top=%d top=%s\n", lua_gettop(L), lua_tostring(L, -1));
if (!n) {
++ n;
return lua_yield(L, 1);
} else {
return 1;
}
}
int main(int argc, char* const argv[]) {
lua_State *L = luaL_newstate();
/* init lua library */
lua_pushcfunction(L, luaopen_base);
if (lua_pcall(L, 0, 0, 0) != 0) {
return 1;
}
lua_pushcfunction(L, luaopen_package);
if (lua_pcall(L, 0, 0, 0 ) != 0) {
return 2;
}
/* create the coroutine */
lua_State *LL = lua_newthread(L);
lua_pushcfunction(LL, func);
lua_pushstring(LL, "hello world");
/* first time resume */
if (lua_resume(LL, 1) == LUA_YIELD) {
printf("first_top=%d top_string=%s\n", lua_gettop(LL), lua_tostring(LL, -1));
/* twice resume */
if (lua_resume(LL, 1) == 0) {
printf("second_top=%d top_string=%s\n", lua_gettop(LL), lua_tostring(LL, -1));
}
}
lua_close(L);
return 0;
}
it core dumps in lua5.1, but works well in lua5.2 if change lua_resume(LL, 1) to lua_resume(LL, NULL, 1).
EDIT: I was actually totally wrong.
You cannot resume a C function.
How do you end a long running Lua script?
I have two threads, one runs the main program and the other controls a user supplied Lua script. I need to kill the thread that's running Lua, but first I need the script to exit.
Is there a way to force a script to exit?
I have read that the suggested approach is to return a Lua exception. However, it's not garanteed that the user's script will ever call an api function ( it could be in a tight busy loop). Further, the user could prevent errors from causing his script to exit by using a pcall.
You could use setjmp and longjump, just like the Lua library does internally. That will get you out of pcalls and stuff just fine without need to continuously error, preventing the script from attempting to handle your bogus errors and still getting you out of execution. (I have no idea how well this plays with threads though.)
#include <stdio.h>
#include <setjmp.h>
#include "lua.h"
#include "lualib.h"
#include "lauxlib.h"
jmp_buf place;
void hook(lua_State* L, lua_Debug *ar)
{
static int countdown = 10;
if (countdown > 0)
{
--countdown;
printf("countdown: %d!\n", countdown);
}
else
{
longjmp(place, 1);
}
}
int main(int argc, const char *argv[])
{
lua_State* L = luaL_newstate();
luaL_openlibs(L);
lua_sethook(L, hook, LUA_MASKCOUNT, 100);
if (setjmp(place) == 0)
luaL_dostring(L, "function test() pcall(test) print 'recursing' end pcall(test)");
lua_close(L);
printf("Done!");
return 0;
}
You could set a variable somewhere in your program and call it something like forceQuitLuaScript. Then, you use a hook, described here to run every n instructions. After n instructions, it'll run your hook which just checks if forceQuitLuaScript is set, and if it is do any clean up you need to do and kill the thread.
Edit: Here's a cheap example of how it could work, only this is single threaded. This is just to illustrate how you might handle pcall and such:
#include <stdlib.h>
#include "lauxlib.h"
void hook(lua_State* L, lua_Debug *ar)
{
static int countdown = 10;
if (countdown > 0)
{
--countdown;
printf("countdown: %d!\n", countdown);
}
else
{
// From now on, as soon as a line is executed, error
// keep erroring until you're script reaches the top
lua_sethook(L, hook, LUA_MASKLINE, 0);
luaL_error(L, "");
}
}
int main(int argc, const char *argv[])
{
lua_State* L = luaL_newstate();
luaL_openlibs(L);
lua_sethook(L, hook, LUA_MASKCOUNT, 100);
// Infinitely recurse into pcalls
luaL_dostring(L, "function test() pcall(test) print 'recursing' end pcall(test)");
lua_close(L);
printf("Done!");
return 0;
}
The way to end a script is to raise an error by calling error. However, if the user has called the script via pcall then this error will be caught.
It seems like you could terminate the thread externally (from your main thread) since the lua script is user supplied and you can't signal it to exit.
If that isn't an option, you could try the debug API. You could use lua_sethook to enable you to regain control assuming you have a way to gracefully terminate your thread in the hook.
I haven't found a way to cleanly kill a thread that is executing a long running lua script without relying on some intervention from the script itself. Here are some approaches I have taken in the past:
If the script is long running it is most likely in some loop. The script can check the value of some global variable on each iteration. By setting this variable from outside of the script you can then terminate the thread.
You can start the thread by using lua_resume. The script can then exit by using yield().
You could provide your own implementation of pcall that checks for a specific type of error. The script could then call error() with a custom error type that your version of pcall could watch for:
function()
local there_is_an_error = do_something()
if (there_is_an_error) then
error({code = 900, msg = "Custom error"})
end
end
possibly useless, but in the lua I use (luaplayer or PGELua), I exit with
os.exit()
or
pge.exit()
If you're using coroutines to start the threads, you could maybe use coroutine.yield() to stop it.
You might wanna take look at
https://github.com/amilamad/preemptive-task-scheduler-for-lua
project. its preemptive scheduler for lua.
It uses a lua_yeild function inside the hook. So you can suspend your lua thread. It also uses longjmp inside but its is much safer.
session:destroy();
Use this single line code on that where you are want to destroy lua script.
lua_KFunction cont(lua_State* L);
int my_yield_with_res(lua_State* L, int res) {
cout << " my_yield_with_res \n" << endl;
return lua_yieldk(L, 0, lua_yield(L, res), cont(L));/* int lua_yieldk(lua_State * L, int res, lua_KContext ctx, lua_KFunction k);
Приостанавливает выполнение сопрограммы(поток). Когда функция C вызывает lua_yieldk, работающая
сопрограмма приостанавливает свое выполнение и вызывает lua_resume, которая начинает возврат данной сопрограммы.
Параметр res - это число значений из стека, которые будут переданы в качестве результатов в lua_resume.
Когда сопрограмма снова возобновит выполнение, Lua вызовет заданную функцию продолжения k для продолжения выполнения
приостановленной C функции(смотрите §4.7). */
};
int hookFunc(lua_State* L, lua_Debug* ar) {
cout << " hookFunc \n" << endl;
return my_yield_with_res(L, 0);// хук./
};
lua_KFunction cont(lua_State* L) {// функция продолжения.
cout << " hooh off \n" << endl;
lua_sethook(L, (lua_Hook)hookFunc, LUA_MASKCOUNT, 0);// отключить хук foo.
return 0;
};
struct Func_resume {
Func_resume(lua_State* L, const char* funcrun, unsigned int Args) : m_L(L), m_funcrun(funcrun), m_Args(Args) {}
//имена функций, кол-во агрументов.
private:
void func_block(lua_State* L, const char* functionName, unsigned int Count, unsigned int m_Args) {
lua_sethook(m_L, (lua_Hook)hookFunc, LUA_MASKCOUNT, Count); //вызов функции с заданной паузой.
if (m_Args == 0) {
lua_getglobal(L, functionName);// получить имя функции.
lua_resume(L, L, m_Args);
}
if (m_Args != 0) {
int size = m_Args + 1;
lua_getglobal(L, functionName);
for (int i = 1; i < size; i++) {
lua_pushvalue(L, i);
}
lua_resume(L, L, m_Args);
}
};
public:
void Update(float dt) {
unsigned int Count = dt * 100.0;// Время работы потока.
func_block(m_L, m_funcrun, Count, m_Args);
};
~Func_resume() {}
private:
lua_State* m_L;
const char* m_funcrun; // имя функции.
unsigned int m_Count;// число итерации.
unsigned int m_Args;
};
const char* LUA = R"(
function main(y)
--print(" func main arg, a = ".. a.." y = ".. y)
for i = 1, y do
print(" func main count = ".. i)
end
end
)";
int main(int argc, char* argv[]) {
lua_State* L = luaL_newstate();/*Функция создает новое Lua состояние. */
luaL_openlibs(L);
luaL_dostring(L, LUA);
//..pushlua(L, 12);
pushlua(L, 32);
//do {
Func_resume func_resume(L, "main", 2);
func_resume.Update(1.7);
lua_close(L);
// } while (LUA_OK != lua_status(L)); // Пока поток не завершен.
return 0;
};