c++ process does not stop - c++

Question

There is something wrong in the code but I do not understand why.
I think one of the reason could be the threads that I used in the code.
The code looks like this.
in main.cpp
vector<thread> t;
vector<future<myClass>> futures;
vector<myClass> chV;
for(int i = 0; i < NumberOfThreads; i++) // NumberOfThreads are 2 here
{
promise<myClass> promises;
futures.push_back(promises.get_future());
t.push_back(thread(MyFunction ,i, PointsNumberInThreads , pointList, std::move(promises)));
}
for_each(t.begin(), t.end(), std::mem_fn(&std::thread::join));
for(int i = 0; i < futures.size(); i++ )
{
// futures.at(i).get().fOut(i); // <-- if I comment out then it gives error. but why?
chV.push_back(futures.at(i).get());
}
myClass c1 = chV.at(0);
myClass c2 = chV.at(1);
cout << "merge start" << endl; // <-- it prints out
c1.Merge(c2);
cout << "merge end" << endl; // <-- does not print out this message. so I have to kill the process
the Merge function in the myClass.cpp I have these sentences at the bottom of the function.
int aSt,aMid,aEnd;
int bSt,bMid,bEnd;
aSt = 0, aEnd = upperV.size(), aMid = (aEnd + aSt)/2;
bSt = 0, bEnd = b.upperV.size(), bMid = (bEnd + bSt)/2;
Point aPoint, bPoint, aPrev, aNext, bPrev, bNext;
aPoint = upperV.at(aMid);
aPrev = upperV.at(aMid-1);
aNext = upperV.at(aMid+1);
bPoint = b.upperV.at(bMid);
bPrev = b.upperV.at(bMid-1);
bNext = b.upperV.at(bMid+1);
bool done = true;
while(done)
{
done = false;
if(orientation(aPoint,bPoint,bPrev) > 0)
{
bEnd = bMid;
bMid = (bEnd + bSt)/2;
bPoint = b.upperV.at(bMid);
done = true;
}
if(orientation(aPoint,bPoint,bNext) > 0)
{
bSt = bMid;
bMid = (bEnd + bSt)/2;
bPoint = b.upperV.at(bMid);
done = true;
}
if(orientation(bPoint,aPoint,aPrev) < 0)
{
aEnd = aMid;
aMid = (aEnd + aSt)/2;
aPoint = upperV.at(aMid);
done = true;
}
if(orientation(bPoint,aPoint,aNext) < 0)
{
aSt = aMid;
aMid = (aEnd + aSt)/2;
aPoint = upperV.at(aMid);
done = true;
}
}
cout << "aPoint = (" << aPoint.x << " , " << aPoint.y << ")" << endl;
cout << "bPoint = (" << bPoint.x << " , " << bPoint.y << ")" << endl;
and they are printed but as I mentioned the process seems to keep running in somewhere.

Answer 1

This is the section of code that I think you should be concerned with. This may
not be why your code is continuing execution as you have stated in your problem, but from the code that you did provide to us; I still think this is an issue that should be addressed.
bool done = true;
while(done)
{
done = false;
if(orientation(aPoint,bPoint,bPrev) > 0)
{
bEnd = bMid;
bMid = (bEnd + bSt)/2;
bPoint = b.upperV.at(bMid);
done = true;
}
if(orientation(aPoint,bPoint,bNext) > 0)
{
bSt = bMid;
bMid = (bEnd + bSt)/2;
bPoint = b.upperV.at(bMid);
done = true;
}
if(orientation(bPoint,aPoint,aPrev) < 0)
{
aEnd = aMid;
aMid = (aEnd + aSt)/2;
aPoint = upperV.at(aMid);
done = true;
}
if(orientation(bPoint,aPoint,aNext) < 0)
{
aSt = aMid;
aMid = (aEnd + aSt)/2;
aPoint = upperV.at(aMid);
done = true;
}
}
You are initially setting this to be true. You are then going into your while loop and are setting it to false. All of your if statements set it back to true. So if any of these conditions are met depending on what orientation() returns then you are not going to break out of this while loop. It may seem to be illogical at first, but I learned over time that a good way to do while loops is in this structure.
bool done = false;
while ( !done ) {
// terminating case - I usually almost always check for this first!
if ( done ) {
break;
}
// Do work here and one or more cases could trip done to be equal to true
// within a if... else if ... else statement or within a series of if satements
// or none of them can, but if they don't then it must be set at the end of the while loop.
// Any one of these or all of them could set done = true and if they don't
// then you should set done = true at the end of the while loop since
// if none of these conditions are met, it will execute the assignment
// before it begins to loop.
if ( some condidition ) {
// Do this
} else if ( some other condidition ) {
// Do this instead
} else {
// Then do this
}
// Or this: Any of these could trip done = true or none of them
// and once again if none of them do, it should be the last thing done
// before the ending brace to your while loop.
if ( this ) {
// Do Work;
}
if ( this ) {
// Do Work;
}
// etc.
// Set this last if none of the if statements trip done to be true;
// otherwise if one or more does trip done, then you can omit this line.
// Note: If none of the if statements trip this flag and you trip it here
// in this section, this loop will occur only one time.
done = true;
}