Java generic function for performing calculations on integer, on double?

Is this possible? Surely if you passed in a double, any 开发者_开发问答sort of function implementation code which casts an object to an Integer would not be able to work unless the cast 'Integer' was specifically used? I have a function like:

public static void increment(Object o){
    Integer one = (Integer)o;
    system.out.println(one++);
}

I cant see how this could be made generic for a double? I tried

public static <E> void increment(E obj){
    E one = (E)obj;
    system.out.println(one++);
}

but it didn't like it?

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In Java, method arguments are passed by value. Therefore, incrementing a method argument will not change the caller's value. So, you must either return the new value:

double increment(double x) {
    return x + 1;
}

or pass a reference type containing the value:

class MutableDouble {
    double value;
}

void increment(MutableDouble d) {
    d.value++;
}

In Java, type arguments must be of reference type, hence double is not a valid value for a type parameter. If you want primitive types, consider overloading the methods instead:

float increment(float f) {
    return f + 1;
}

double increment(double d) {
    return d + 1;
}

If you really want generics, you can do:

@SuppressWarnings("unchecked")
<N extends Number> N increment(N n) {
    if (n instanceof Double) {
        return (N) (Double) (((Double)n) + 1);
    } else if (n instanceof Float) {
        return (N) (Float) (((Float)n) + 1);
    } else if ( ...
        // handle remaining cases
    } else if (n == null) {
        throw new NullPointerException();
    } else {
        throw new IllegalArgumentException("Unexpected number type: " + n.getClass());
    }
}

This is godawful ugly though. What are you trying to accomplish by defining such a method? There's probably is an easier way ...

---

You may want to read up on semantics that higher-level specialized languages have defined from cross-type arithmetic. A good example that I recommend is JSP EL specification.

It basically spells out how to implement such function...

http://www.google.com/url?sa=t&source=web&cd=7&ved=0CEwQFjAG&url=http%3A%2F%2Fflaka.googlecode.com%2Ffiles%2Fjsp-2_1-fr-spec-el.pdf&rct=j&q=jsp%20el%20specification&ei=erkwTZrZGJHogQfbxMWiCw&usg=AFQjCNH6oPO7WwgiHCBkST5vgsBxMZmWaQ&sig2=0hRqxe6G0brRGuzNOxEhGw&cad=rja

---

What if you downcast it to a Double?

public static void increment(Object o) {
    Double d= (Double) o;
    System.out.println(d + 1.0);
}

Update: I tried this and it worked:

public static void increment(Object o) {
    Double d = Double.parseDouble(o.toString());
    System.out.println(d + 1.0);
}

---

public static double increment(double d)
{
  return d+1;
}

That will work for all primitive value types. Then all you have to do is cast the result ;-)

---

Assuming that you want to increment and get new value (not change immutable Integer instance), you can do something like this:

interface Arithmetics<T> {
    T inc(T val);
}

class IntegerArithmetics implements Arithmetics<Integer> {
    Integer inc(Integer val) { return val+1; }
}

//similarly DoubleArithmetics, BigIntegerArithmetics, ...

And call it to get your method:

public static <E> void increment(E obj, Arithmetics<E> arit){
    System.out.println(arit.inc(obj));
}

Generic Java Math library does exactly that for you.

Only thing is that you need to pass Arithmetics object around. If that is a big problem, you can initialize a Map<Class,Arithmetics>, and use map.get(obj.getClass()) to get corresponding Arithmetics.