Force Number to parse integer String as Double in Java

Question

How can I parse a set of numeric strings into a generic array? My whole problem can be reduced to the following:

static <T extends Number> T[] parseBulk(String bulk, Class<T> type) throws ParseException {
    NumberFormat numberFormat = NumberFormat.getInstance();
    numberFormat.setParseIntegerOnly(false);

    String[] elements = bulk.split(";");
    T[] result = (T[])Array.newInstance(type, elements.length);
    for (int i= 0; i< elements.length; i++)
        result[i] = (T)numberFormat.parse(itemElements[i]); // here it crashes
    //  result[i] = type.cast(numberFormat.parseObject(itemElements[i])); // does not work either
    return result;
}

...

Double[] shouldBeDoubles = parseBulk("15.5;10", Double.class);

Looks like Number determines resulting type by content of the String given, which yields a Double for "15.5" and a Long for "10".

Then it happily crashes trying to convert Long to T, which is Double.

Thank you,

Update : added numberFormat to the source code. Excuse me :)

Answer 1

NumberFormat.parse(String source) returns a Number . The runtime type of the returned instance can be any sub-class of Number , so if it returns a Long , you can't cast it to Double , as you learned.

If you want to force the numeric type that the String s will be converted to, you can pass a parser Function to the method:

static <T extends Number> T[] parseBulk(String bulk, Class<T> type, Function<String,T> parser) {
    String[] elements = bulk.split(";");
    T[] result = (T[])Array.newInstance(type, elements.length);
    for (int i= 0; i< elements.length; i++) {
        result[i] = parser.apply(elements[i]);
    }
    return result;
}

And call it, for example, with:

Double[] shouldBeDoubles = parseBulk("15.5;10", Double.class,Double::valueOf);

This will output:

[15.5, 10.0]