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

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.

use Long.parseLong(str); instead of Long.getLong(str); – user370305 Sep 11 '11 at 07:31 — user370305, Sep 11 '11 at 07:31

MeLight · Accepted Answer · 2011-09-11T18:36:51.913

122

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).

edited Sep 11 '11 at 18:36

answered Sep 11 '11 at 08:28

MeLight

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10

I have often wondered what `getLong` and the other getX of the primitive types where supposed to do. It seems to me to be a very poor design choice to have `getLong` on the `Long` class which is related to system variables or perhaps just a curse to those of us who, like my self, use ctrl+space too extensively. Great response. – abondoa Mar 20 '14 at 15:41

score 12 · Answer 2 · edited Oct 06 '15 at 13:58

12

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

edited Oct 06 '15 at 13:58

Andrea

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answered Sep 11 '11 at 07:20

confucius

13,127
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score 7 · Answer 3 · answered Sep 11 '11 at 07:24

7

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

answered Sep 11 '11 at 07:24

Android Killer

18,174
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score 7 · Answer 4 · answered Sep 11 '11 at 07:53

7

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

answered Sep 11 '11 at 07:53

Y.A.P.

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score 6 · Answer 5 · answered May 12 '16 at 18:22

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.

score 2 · Answer 6 · answered Aug 23 '22 at 14:52

2

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

answered Aug 23 '22 at 14:52

aarush gandhi73

79
6

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

6 Answers6

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

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