How can I make a overhead a method with the parameter HashMap?

Question

I have two methods that make the same thing, but one receives a HashMap<Integer,double[]> and the other receive a HashMap<Integer,int[]>. How can I make this without to duplicate code.

You could convert to `Map`, but that would involve changing callers of the method and Object overhead. Guess, there's no better way, though. — qqilihq, Oct 11 '14 at 19:56
The code you posted won't even compile, you need to use `Integer[]` instead of `int[]` — msrd0, Oct 11 '14 at 19:57
@msrd0 a primitive array is an `Object`, and thus eligible as a type argument. It compiles just fine. — Mureinik, Oct 11 '14 at 19:59
@qqilihq Because Java doesn't allow types like `int` or `double` to be templates. You always need the wrapper classes like `Integer` or `Double` — msrd0, Oct 11 '14 at 19:59

score 0 · Answer 1 · answered Oct 11 '14 at 19:59

0

Make two methods, one contains the real code, acception the version with the double-array and the other method, accepting the integer-version just calls this method but first converts the integer-array to a double array.

public void doSomething(HashMap<Integer, int[]> arg0) {

   //convert int[] to double[] and fill this HashMap
   HashMap<Integer, double[]> converted;

   doSomething(converted);
}

public void doSomething(HashMap<Integer, double[]> arg0) {

   //your logic here

}

answered Oct 11 '14 at 19:59

rogi1609

438
3
8

this won't work because of erasure both become `doSomething(HashMap – Oct 11 '14 at 20:06
yes, noticed after posting. thank you for alerting me to that fact. – rogi1609 Oct 11 '14 at 20:10
@JarrodRoberson Could I to make this with generics types ? – fmassica Oct 11 '14 at 20:32
this will never work!`doStuff(HashMap) and doStuff(HashMap) have the same erasure` – Oct 11 '14 at 21:11

How can I make a overhead a method with the parameter HashMap?

1 Answers1