Question

Need urgent help on thread: the goal here is the separtemask will take each image and separate different contours and for each contour in the image it will call handleobject thread. So every for loop will call the handeobject thread. However, object index variable needs to be passed in each thread. But only last value of objectndex is passed, this is becuase the speratemask function loops and repalces the value of obj.objindx and only the last value of obj.objindx is
passed to all the threads. Is there anyway to pass each objectindex
value in handleobject. The code runs fine if we uncomment the pthread_join(tid[objectIndex],NULL); but it will not give a parralel program

void separateMask(IplImage *maskImg)
{    
  for(r = contours; r != NULL; r = r->h_next)
  {
    cvSet(objectMaskImg, cvScalarAll(0), NULL);
    CvScalar externalColor = cvScalarAll(0xff);
    CvScalar holeColor = cvScalarAll(0x00);
    int maxLevel = -1; 
    int thinkness = CV_FILLED; 
    int lineType = 8; /* 8-connected */
    cvDrawContours(objectMaskImg, r, externalColor, holeColor, maxLevel, thinkness,lineType, cvPoint(0,0));;
    obj.objectMaskImg1[objectIndex]=(IplImage *) malloc(sizeof(IplImage));
    obj.objectMaskImg1[objectIndex]=objectMaskImg;
    obj.objindx=objectIndex;
    obj.intensityOut1=intensityOut;
    obj.tasOut1=tasOut;
    pthread_create(&tid[objectIndex],NULL,handleObject,(void *)&obj);
    //pthread_join(tid[objectIndex],NULL);
    printf("objectindx %d\n",obj.objindx);
    objectIndex++;

  }
  // cvReleaseImage(&objectMaskImg);
  //cvReleaseMemStorage(&storage);
  printf("Exitng Separatemask\n");

}


void* handleObject(void *arg)
{
  int i, j;
  handle *hndl;
  hndl=(handle *) malloc(sizeof(handle));
  hndl=(handle*)arg;
  pthread_mutex_t lock=PTHREAD_MUTEX_INITIALIZER;
  pthread_mutex_lock(&lock);
  IplImage *pImg;
  float statistics_ratio[3][9];
  pthread_t tid3;
  tas3 tas2;
  pImg = cvLoadImage("image.tif", CV_LOAD_IMAGE_ANYCOLOR | CV_LOAD_IMAGE_ANYDEPTH);
  if(pImg == NULL)
  {
    fprintf(stderr, "Fail to load image %s\n", "tiff file");
    return ;
  }
  tas2.pImg1=pImg;
  printf("tst%d\n",hndl->objindx);
  tas2.x=hndl->objindx;
  tas2.objectMaskImg1=hndl->objectMaskImg1[tas2.x];
  tas2.statistics_ratio[3][9]=statistics_ratio[3][9];
  double mean = average_intensity(pImg, tas2.objectMaskImg1); 
  int total = total_white(pImg, tas2.objectMaskImg1);
  pthread_mutex_unlock(&lock);

  printf("Exiting handle object thread_id %d\n\n", pthread_self());
}

Answer 1

This function appears to have issues

void* handleObject(void *arg)

Firstly

    pthread_mutex_t lock=PTHREAD_MUTEX_INITIALIZER;
    pthread_mutex_lock(&lock);

this is a locally created mutex – but created WITHIN the thread function. So you lock it but since nothing else can see the mutex, why do you need it??? It gives no synchronization functionality if no other threads can see it.
Secondly

    float statistics_ratio[3][9];
    pthread_t tid3;
    tas3 tas2;
    pImg = cvLoadImage("image.tif", CV_LOAD_IMAGE_ANYCOLOR | CV_LOAD_IMAGE_ANYDEPTH);
    if(pImg == NULL){
        fprintf(stderr, "Fail to load image %s\n", "tiff file");
        return ;
    }
    tas2.pImg1=pImg;
    printf("tst%d\n",hndl->objindx);
    tas2.x=hndl->objindx;
    tas2.objectMaskImg1=hndl->objectMaskImg1[tas2.x];
    tas2.statistics_ratio[3][9]=statistics_ratio[3][9];

you create a local uninitialised 2d float array statistics_ratio, do nothing with it then assign it to another locally created object member. This appears to be meaningless, as does the declaration of another pthread instance tid3.

It doesn’t really matter since nothing else can see the thread but you return from inside this function if pImg == NULL without first unlocking the mutex.

It is very hard to see why your code doesnt work or what it is meant to do, but perhaps the things highlighted above may help. You are creating a lot of local variables within your thread functions which are not being used. I am not sure if you need some of these to be global instead – particularly the mutex (if indeed you need one at all).