Threading in QB64pe (again)

[justsomeguy]

Hello all,

I've been revisiting my earlier attempt at threading in QB64pe. You can look at the old thread at https://qb64forum.alephc.xyz/index.php?topic=3865.0 Thanks to guys that responded to that thread and helped me. 

I'm happy to say I made a bit of progress. I managed to get two separate free running threads to run concurrent with my main QB64 program. They don't do much, but they show its possible.

In order to get it working I had to cheat and use a c header file to make a wrapper for pthreads. This makes the declarations easier.

Code: (Select All)

// pthreadGFXTest.h
// Threading Header
#include "pthread.h"
// Only needed for the SIGTERM Constant
#include <signal.h>
// Initialize Threads
pthread_t thread0;
pthread_t thread1;
// Easy way to determine if a thread is running
bool threadRunning0 = false;
bool threadRunning1 = false;
// Setup Mutexes for each of the threads.
static pthread_mutex_t mutex0;
static pthread_mutex_t mutex1;
// QB's names for the threaded Subs
// You can locate these in your ''qb64pe/internal/temp'' folder.
// I found these in the 'main.txt'
void SUB_LINES();
void SUB_CIRCLES();
// wrap the subs so that you can easily get the void* for pthread
void* RunLines(void *arg){
    SUB_LINES();
}
void* RunCircles(void *arg){
    SUB_CIRCLES();
}
// These are the commands that are accessed by you program
void invokeLines(){
    if (!threadRunning0) {
        int iret = pthread_create( &thread0, NULL, RunLines, NULL);
        pthread_mutex_init(&mutex0, NULL);
        threadRunning0 = true;
    }
}
void invokeCircles(){
    if (!threadRunning1) {
        int iret = pthread_create( &thread1, NULL, RunCircles, NULL);
        pthread_mutex_init(&mutex1, NULL);
        threadRunning1 = true;
    }
}
void joinThread0(){
    pthread_join(thread0,NULL);
    threadRunning0 = false;
}
void joinThread1(){
    pthread_join(thread1,NULL);
    threadRunning1 = false;
}
void exitThread(){
    pthread_exit(NULL);
}
void killThread0(){
    if (threadRunning0) {
        int iret = pthread_kill(thread0, SIGTERM);
    }
}
void killThread1(){
    if (threadRunning1) {
        int iret = pthread_kill(thread1, SIGTERM);
    }
}
void lockThread0(){
    pthread_mutex_lock(&mutex0);
}
void unlockThread0(){
    pthread_mutex_unlock(&mutex0);
}
void lockThread1(){
    pthread_mutex_lock(&mutex1);
}
void unlockThread1(){
    pthread_mutex_unlock(&mutex1);
}

The test program draws lines in one thread and circles in another thread. You start and stop the threads by pressing '1' and '2' and 'ESC' quits.

Code: (Select All)

'***********************************************************************************

' Proof of concept threading in QB64pe.

' by justsomeguy

'***********************************************************************************



' Thread Library Declaration

DECLARE LIBRARY "./pthreadGFXTest"

  SUB invokeLines ' start Lines thread

  SUB invokeCircles ' start Circles thread

  SUB joinThread0 ' wait til thread is finished

  SUB joinThread1 ' wait til thread is finished

  SUB exitThread ' must be called as thread exits

  SUB killThread0 ' kill the thread

  SUB killThread1 ' kill the thread

  SUB lockThread0 ' mutex lock

  SUB unlockThread0 ' mutex unlock

  SUB lockThread1 ' mutex lock

  SUB unlockThread1 ' mutex unlock

END DECLARE



' Global variables

DIM SHARED AS INTEGER q0, q1 ' quit signals

DIM AS STRING ky



' Setup screen

_TITLE "Thread test"

SCREEN _NEWIMAGE(1024, 768, 32)

_FONT 8

CLS



' Fire up freerunning threads

invokeCircles

invokeLines



' Campout in an infinite loop

DO

  ky = INKEY$

  LOCATE 1, 1

  PRINT "Lines are drawn on one thread, Circles are drawn in a second thread."

  PRINT "Press '1' to toggle the Line drawing thread. "

  PRINT "Press '2' to toggle the Circle drawing thread."

  PRINT "Press 'ESC' to exit."



  IF ky = "1" THEN

    q0 = NOT q0

    IF q0 THEN

      joinThread0

    ELSE

      invokeLines

    END IF

  END IF



  IF ky = "2" THEN

    q1 = NOT q1

    IF q1 THEN

      joinThread1

    ELSE

      invokeCircles

    END IF

  END IF



  ' Quit the whole program

  IF ky = CHR$(27) THEN q0 = -1: q1 = -1: joinThread0: joinThread1: SYSTEM

LOOP





'***********************************************************************************

' Threaded

'***********************************************************************************



SUB lines ()

  ' Free running loop

  DO

    ' lock a mutex, just to be safe

    lockThread0

    ' Do something

    LINE (RND * _WIDTH, RND * _HEIGHT)-(RND * _WIDTH, RND * _HEIGHT), _RGB32(RND * 255, RND * 255, RND * 255)

    ' unlock mutex

    unlockThread0

    ' do I need to jump out?

  LOOP UNTIL q0 = -1

  ' Must call exitThread when leaving, so that joinThread works.

  exitThread

END SUB



SUB circles ()

  ' Free running loop

  DO

    ' lock a mutex, just to be safe

    lockThread1

    ' Do something

    CIRCLE (RND * _WIDTH, RND * _HEIGHT), RND * 50, _RGB32(RND * 255, RND * 255, RND * 255)

    ' unlock mutex

    unlockThread1

    ' do I need to jump out?

  LOOP UNTIL q1 = -1

  ' Must call exitThread when leaving, so that joinThread works.

  exitThread

END SUB

My goal, is to get my 2d physics engine to reside in a separate thread and have it free running computing collisions and motion while the main thread handles I/O and other logic.

I'm using QB64pe 3.11.0. I tested this on Linux Mint, MacOS and Windows 10. On windows I had to add '-pthread' to compiler settings and '--static' to the linker settings.



To get this running on your computer, copy the header to your favorite text editor and save the file under 'pthreadGFXTest.h' Then copy the source to the same directory as the header and make sure your compiler and linker settings are correct. I'm not sure if its necessary, but I save my EXE to the source folder.

Beware that if you decide try playing around with the code, that it could crash in some wild ways. Error messages will not make sense, and it might run a bit and lockup for unknown reasons. QB64pe is not meant to be run like this, so there will not be much help if you try.