如何在Java中组织3个嵌套循环？

Question

Here is my loops structure.

int k = 0;
do {
    ....                        

    do {
        fc = Recognize();
        ....
        do {
            ....
        } while(fc != false);

        k++;
    } while(k != 20);

    if (k == 20) {
        return;
    }
} while(true);

I need to end all loops when either k equals 20 or fc equals true.
I tried different ways to do this but failed. The code above is my last attempt. Need help in fixing that.

   grabber.start();
   int k=0;
   do {   
     grabber.stop();
     grabber.start();

     img = grabber.grab();
     if (img != null) {
       canvas.showImage(img);
       canvas.pack();
       cvWaitKey(0);
     }
     do {  
       fc=  Recognize();
       grabber.start();
       do {
         img = grabber.grab();
         if (img != null) {
            CvHaarClassifierCascade cascade = new 
            CvHaarClassifierCascade(cvLoad(XML_FILE));
            CvMemStorage storage = CvMemStorage.create();
            CvSeq sign = cvHaarDetectObjects(img,
              cascade,storage, 1.5, 3,CV_HAAR_DO_CANNY_PRUNING);

            cvClearMemStorage(storage);
            total_Faces = sign.total();     

            for(int i = 0; i < total_Faces; i++){
              CvRect r = new CvRect(cvGetSeqElem(sign, i));
              cvRectangle (img,
                cvPoint(r.x(),r.y()),
                cvPoint(r.width()+r.x(),r.height()+ r.y()),
                CvScalar.RED,2,CV_AA,0);

                x=r.x();
                y=r.y();
                h=r.height();
                w=r.width();
              }         

              cvFlip(img, img, 1);
              canvas.showImage(img);
              canvas.pack();
              cvWaitKey(0);
           }
         } while(fc!=false);

         k++;    
       }  while(k!=21);  
     } while(true);  
  }

Answer 1

Short answer: don't do that.

Check out Clean code by Robert Martin; and understand & apply the single layer of abstraction principle.

There are good reasons why various tools measure "code complexity" by especially looking for such kind of nested structures.

I understand that many people with a more "mathematical" background think that it is OK when computer programs resemble "mathematical" structures. But well: that is wrong. Computer programs are written ... to be read by humans. When you read a piece of source code, the main thing that counts is: how long does it take you to understand what is going on.

I guarantee you, no matter how you structure a loop in a loop in a loop; in one week from now you will have major difficulties to understand what is going on. Therefore it is worth to sit down for an hour or more to come up with a better solution. Unfortunately, your example doesn't show the real problem you are trying to solve; thus it is a hard to give concrete tips on how to do that in your case.

Answer 2

while(k!=21 && fc){...} is enough. Or do {...} while(k!=21 && fc); if need be. The outermost loop is useless as well as some of the if s in the middle.

Answer 3

The only loop that needs to be written is

int k = 0;
do {
      fc = Recognize();
      k++  ;
   } while(k!=20 || fc );