CODEWITHSUNDEEP
javaconstructorargumentsdouble
flixe
flixe

Reputation: 646

Java: constructor argument (type double) always 0

i have a class with an construktor with two arguments of type double.

public class NetcdfHandler {

double x;
double y;

public NetcdfHandler (double x, double y){
    x = this.x;
    y = this.y;
}
}

When i call this in the doGet Method of my Servlet, the given double values somehow are not "received" by the constructor. That means after creating an instance of NetcdfHandler, x and y both have the value 0.0 (but not "null") although "Lat" and "Lng" are set up correctly:

public class Test extends HttpServlet {
    private static final long serialVersionUID = 1L;

    double Lat;
    double Lng;

public Test() {
        super();
    }

protected void doGet(HttpServletRequest req, HttpServletResponse rsp) throws ServletException, IOException {
        Lat = Double.parseDouble(req.getParameter("lat"));
        Lng = Double.parseDouble(req.getParameter("lng"));

        NetcdfHandler nc = new NetcdfHandler(Lat, Lng);
}

I guess it's quite a beginners mistake, but i couldn't figure out whats the problem here. Maybe someone can help?

Upvotes: 0

Views: 140

Answers (1)

Tagir Valeev
Tagir Valeev

Reputation: 100199

Your constructor is incorrect. Write this way:

public NetcdfHandler (double x, double y){
    this.x = x;
    this.y = y;
}

Currently you're doing the opposite thing: put the default field values (which are 0.0) into parameters. So fields are not changed and parameters are forgotten upon constructor exit.

If you are not going to change fields x and y after construction, it's better to declare them final:

final double x;
final double y;

This way it's easier to prevent some programming mistakes. In particular your code would result in compilation error like "final field is not initialized in constructor".

Upvotes: 4

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