Question

I am using some code with generics in which I am unsure why there is a compile error being thrown.

  public static void main(String[] args) {

    System.out.print("s");
    setInteger(prepareNumber("1")); // Error here! solved by casting: setInteger((Integer)prepareNumber("1"))
    Integer c = prepareInteger("1");
    System.out.print(c);

  }

  public static Integer prepareInteger(Object number) {
    return prepareNumber(number).intValue();
  }

  private static <T extends Number> T prepareNumber(Object number) {

    T returnValue = null;

    // handle a blank number
    if (number == null || !NumberUtils.isNumber(number.toString())) {
      returnValue = null;
    } else {
      if (number.toString().contains(".")) {
        returnValue = (T) Double.valueOf(number.toString());
      } else {
        returnValue = (T) Integer.valueOf(number.toString());
      }
    }

    return returnValue;
  }

  private static void setInteger(Integer a){    
  }

Fixing this is easy by just doing a casting, but that is undesired, as the generics should detect the method needs an Integer and should provide the return value as such and not as the super type.

PS: Do not mind the parsing code, is just testing stuff.

Thanks!

Answer 1

The compiler is unable to infer what type prepareNumber("1") is supposed to return in this case. You can get it to infer correctly by placing the result in a temporary variable:

Integer value = prepareNumber("1");        
setInteger(value);

Or you if parseNumber is a static method in class YourClass you can explicitly tell the compiler what type you expect like so:

setInteger(YourClass.<Integer>prepareNumber("1"));

---

Or if you are okay with changing the method signature to help the compiler:

setInteger(prepareNumber("1", Integer.class));

private static <T extends Number> T prepareNumber(Object number, Class<T> classType) {
   ....
}