Long.getLong() failing, returning null to valid st

Question

I've spent the past two hours debugging what seems extremely unlikely. I've stripped the method of a secondary Android Activity to exactly this:

public void onClick(View v) {
        String str = "25";
        long my_long = Long.getLong(str);
} // onClick (v)

And yeah, I get a crash with the good ol' NullPointerException:

09-11 02:02:50.444: ERROR/AndroidRuntime(1588): Uncaught handler: thread main exiting due to uncaught exception 09-11 02:02:50.464: ERROR/AndroidRuntime(1588): java.lang.NullPointerException

It looks like (from other tests) that Long.getLong(str) returns NULL, which is driving me bonkers. WHAT AM I MISSING?

Thanks in advance. I'm okay with stupidly missing the obvious, but my sanity is on the line.

Answer 1

You are missing the fact that Long.getLong(String str) is not supposed to parse a String to a long, but rather to return a long value of a system property represented by that string. As others have suggested, what you actually need is Long.parseLong(String str).

Answer 2

I think you are using wrong function use Long.parseLong(str) then you can get the right answer.

Answer 3

You can use Long.parseLong(String), instead of getLong(String): it will solve the problem.

Answer 4

To understand this, some examples:

Long val= Long.getLong("32340");

returns: null

Long val = Long.getLong("32340", 3000);

returns: 3000

Using Long.parseLong() :

Long val  = Long.parseLong("32340");

returns: 32340

The documentation describe getLong() method as :

Returns the Long value of the system property identified by string.

this is the code of the getLong() method and only get a property value defined by a string:

  public static Long getLong(String string) {
        if (string == null || string.length() == 0) {
            return null;
        }
        String prop = System.getProperty(string);
        if (prop == null) {
            return null;
        }
        try {
            return decode(prop);
        } catch (NumberFormatException ex) {
            return null;
        }
    }

If you want to parse a String to Long, the best way is using Long.parseLong() method.

Answer 5

Long.parseLong(someString) approved. Don't forget to catch NumberFormatException if there's a probability of unparsable string.