Question

I have a double var

public double votes(){
    double votexp = 0;

      for(Elettore e:docenti.values()){
        if(e.getVoto()==true)          //everytime this is true increment by 1
        {
            votexp+=1.0;
        }   
    }
    for(Elettore e:studenti.values()){
        if(e.getVoto()==true)         //everytime this is true increment by 0.2
        {
            votexp+=0.2;
        }
    }
    for(Elettore e:pta.values()){
        if(e.getVoto()==true)        //everytime this is true increment by 0.2
        {
            votexp+=0.2;
        }
    }
    return votexp;
}

In my case the variable shoud be incremented to 2.6 but votexp returns 2.6000000000000005
how can i fix this by using the same double var and return a double precision number ?

Answer 1

You are accumulating a rounding error. The simplest thing to do is to use an long (or int and only use a double at the end. (Or a BigDecimal and double at the end, but this is overly complicated)

public double votes() {
    long votexp = 0;
    for (Elettore e : docenti.values())
        if (e.getVoto())          //everytime this is true increment by 1
            votexp += 10;
    for (Elettore e : studenti.values())
        if (e.getVoto())         //everytime this is true increment by 0.2
            votexp += 2;
    for (Elettore e : pta.values())
        if (e.getVoto())        //everytime this is true increment by 0.2
            votexp += 2;
    return votexp / 10.0;
}

---

double a = 26 / 10.0;
System.out.println(a);

prints

2.6

As the Double.toString() “knows” values can be imprecise, it will do a small amount of rounding. This rounding is limited so the result of operating on two double can have an error too large to hide. If you round your last result correctly, the error will be small enough it won’t cause a problem.