Why does this Java code throw a NullPointerException?

Question

Why does the following Java code throw a NullPointerException?

public static void main(String[] args) {
    getInteger(null);
}

public static Integer getInteger(Number n) {
    return (n == null || n instanceof Integer) ? (Integer)n : n.intValue();
}

EDIT: I added parentheses to end the confusion about "whether I am sometimes returning a boolean".

@TimBiegeleisen That's what I thought too. Test it yourself :) — Hans Brende, Dec 03 '15 at 07:04
as a sidenode, `n instanceof Integer` will also fail if `n` is null, so you are basicly checking it twice. — SomeJavaGuy, Dec 03 '15 at 07:09
if LHS condition made true then how RHS side goes compiter , if i am not wrong then LHS return true then it won't go to RHS, Question is very good, i appreciate, but i am looking that many answer came and also get upvote , but no body explaining my this query. can any one please and make me correct — Vishal Gajera, Dec 03 '15 at 07:14
@TimBiegeleisen Nope. Try the code with parentheses. Still throws NullPointerException. — Hans Brende, Dec 03 '15 at 07:19
@HansBrende change `n.intValue()` to `Integer.valueOf(n.intValue())` (all the current answers seem incorrect). — assylias, Dec 03 '15 at 07:25
@assylias thank you for the only correct response! Java auto-unboxing at work... shoulda known. — Hans Brende, Dec 03 '15 at 07:30
@assylias Can you explain what is actually happening? In the case of a `null` input, what is the boxed type of `n` that is throwing an exception with the case `(Integer)n`? This is not clear to me. Please post an answer. — Tim Biegeleisen, Dec 03 '15 at 07:32
@Tim, since the second part of the expression is `n.intValue()` which gives an **primitive int**, the compiler uses unboxing on the first part as well... So `(Integer) n` becomes `((Integer) n).intValue()` in compiled code. So this throws a NPE on null input. Hope it clears. — Codebender, Dec 03 '15 at 07:34

score 4 · Answer 1 · edited Dec 03 '15 at 11:41

4

Credit is completely due to @assylias who discovered the answer. Use this code instead:

public static Integer getInteger(Number n) {
    return (n == null || n instanceof Integer) ? (Integer)n : Integer.valueOf(n.intValue());
}

Because of weird boxing/unboxing rules, since n.intValue() returns a primitive int, the compiler was unboxing n in the expression (Integer)n to a primitive as well. ~~And null cannot be assigned to a primitive (just try doing that in your IDE).~~

Edit:

The NPE is not really because null is being assigned to a primitive (the compiler would not do that), but it's because unboxing is done by calling Integer.intValue() method and the method is called on a null reference in this case.

edited Dec 03 '15 at 11:41

Codebender

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answered Dec 03 '15 at 07:07

Tim Biegeleisen

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But this doesn't tell me why my code failed. – Hans Brende Dec 03 '15 at 07:08
Yes it does. In the case of `null`, you were trying to return a `boolean` value in the RHS expression. – Tim Biegeleisen Dec 03 '15 at 07:09
3

@TimBiegeleisen don't you think that IDE should alert the user that this statement might return boolean? I tried the code and I was thinking that.. – Danyal Sandeelo Dec 03 '15 at 07:10
@DanyalSandeelo Very good question. – Tim Biegeleisen Dec 03 '15 at 07:11
Yes... It should ve been a compiler error if he is returning a boolean value instead of Integer... The catch is somewhere else. – Codebender Dec 03 '15 at 07:11
if LHS condition made true then how RHS side goes compiter , if i am not wrong then LHS return true then it won't go to RHS – Vishal Gajera Dec 03 '15 at 07:14
@TimBiegeleisen I edited my question to add parentheses to remove the confusion about returning a boolean. – Hans Brende Dec 03 '15 at 07:17
@TimBiegeleisen but `n instanceof Integer` has an inbuild `null` check and does both, check for class and check if it´s not equal to null. The Exception actually happens at `(Integer)n`, just in combination with the ternary operator. – SomeJavaGuy Dec 03 '15 at 07:27
@assylias I see it now. But if both conditions are being boxed into `Integer`, why would there be a problem? – Tim Biegeleisen Dec 03 '15 at 07:30
1

@TimBiegeleisen the problem is that both are being UNboxed to int. – Hans Brende Dec 03 '15 at 07:34
@TimBiegeleisen looks like someone edited your answer incorrectly :/ – Hans Brende Mar 19 '19 at 17:46

score 0 · Answer 2 · edited Dec 03 '15 at 07:17

0

Because here:

return n == null || n instanceof Integer ? (Integer)n : n.intValue();
                                             here

You are trying to convert the n (which is null) to Integer.

edited Dec 03 '15 at 07:17

ΦXocę 웃 Пepeúpa ツ

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answered Dec 03 '15 at 07:06

KostasC

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But null CAN be converted to an Integer. – Hans Brende Dec 03 '15 at 07:07
I tried the code, the check is false so the code is not going there, – Danyal Sandeelo Dec 03 '15 at 07:07
@KostasC all know because of casing time throw NullPointerException, but see n== null returns true then how moves to right side of short-circulate OR operator – Vishal Gajera Dec 03 '15 at 07:10
while using || left side return true then it want more to right side – Vishal Gajera Dec 03 '15 at 07:10
if LHS condition made true then how RHS side goes compiter , if i am not wrong then LHS return true then it won't go to RHS – Vishal Gajera Dec 03 '15 at 07:13
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This answer is incorrect... See duplicate. – assylias Dec 03 '15 at 07:38
1

@assylias Thanks for correcting us. – Gaurav Jeswani Dec 03 '15 at 09:06

Why does this Java code throw a NullPointerException?

2 Answers2

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