Why does the Java increment operator allow narrowing operations without explicit cast?

Question

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Java += operator

In Java, this is not valid (doesn't compile), as expected:

long lng = 0xffffffffffffL;
int i;
i = 5 + lng;    //"error: possible loss of magnitude"

But this is perfectly fine (?!)

long lng = 0xffffffffffffL;
int i = 5;
i += lng;       //compiles just fine

This is obviously a narrowing operation, that can possibly exceed the int range. So why doesn't the compiler complain?

This question invites discussion instead of a clear-cut answer. — Marko Topolnik, Dec 19 '12 at 10:08

score 14 · Answer 1 · answered Dec 19 '12 at 10:06

14

This is defined in the JLS #15.26.2:

A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T) ((E1) op (E2)), where T is the type of E1, except that E1 is evaluated only once.

In other words, i += lng performs a cast implicitly.

answered Dec 19 '12 at 10:06

assylias

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not jelous, but i said the same darn thing in my answer a minute befor you and i got no reps.. :P – PermGenError Dec 19 '12 at 10:13
2

@GanGnaMStYleOverFlowErroR You just rephrased OP's question. assylias got upvoted for referencing the JLS. As opposed to you, he posted that in the **first** revision, and was the first one to post it; you needed about 5 revisions to get there (and only after seeing others' answers). – Marko Topolnik Dec 19 '12 at 10:14
@GanGnaMStYleOverFlowErroR Sorry about that - but I'm certainly not going to give away the 3 hats I just got ;-) – assylias Dec 19 '12 at 10:15
dammit, is it all about the hats now?? :) – Marko Topolnik Dec 19 '12 at 10:16
@MarkoTopolnik Yes - reputation is way overstated. Only hats count. – assylias Dec 19 '12 at 10:17
@MarkoTopolnik i did reference JLS in my answer too .. but whatever, i am not really greedy abouts reps or hats :P – PermGenError Dec 19 '12 at 10:19
@assylias haha, why would you say sorry .. i was just makin a conversation thats all.. +1 anyways to have more reps than the question which is already asked :) – PermGenError Dec 19 '12 at 10:19
@GanGnaMStYleOverFlowErroR ...in your first revision of the answer, or in sixth? I know which, I've been watching your edits. – Marko Topolnik Dec 19 '12 at 10:21
@MarkoTopolnik hahah, yeah it was later on i referenced the JLS, but still my answer still reflects what's said in JLS doesn't it ?? lemme know if there's something wrong in my answer :) – PermGenError Dec 19 '12 at 10:24
@GanGnaMStYleOverFlowErroR If you want to know why you didn't get the upvotes, it is because your answer was sitting voteless at the moment assylias posted his; everybody upvoted the clean answer backed with a JLS reference and from that point on it was irrelevant to everybody that there happened to be other zero-score answers below that said exactly the same thing. Learn from this and become a better contributor to StackOverflow. **Especially** because soon you'll be a *k-guy*. – Marko Topolnik Dec 19 '12 at 10:27
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@MarkoTopolnik mark as said earlier, i am not greedy or something. i totally agree with you that my answer wasn't as clear as assylias or NPE, i was only making a conversation :p, and i would love to be a better cobntributor to SO .. :) – PermGenError Dec 19 '12 at 10:30

PermGenError · Accepted Answer · 2012-12-19T10:10:43.000

5

i += lng; compound assignment operator cast's implicitly.

i+=lng; 
is same as 
i = int(i+lng);

FROM JLS:

A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T) ((E1) op (E2)), where T is the type of E1, except that E1 is evaluated only once.

edited Dec 19 '12 at 10:10

answered Dec 19 '12 at 10:05

PermGenError

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So it seems, but why? It seems to me to go against the gist of the language. – Cristian Diaconescu Dec 19 '12 at 10:06
@CristiDiaconescu, This implicit cast happens without errors because there is no syntax available to make it explicit. For example you wouldn't be able to `*=1.5` an integer otherwise. – Robert Nov 13 '14 at 20:19

score 3 · Answer 3 · answered Dec 19 '12 at 10:07

The compiler does not complain because, according to JLS §15.26.2. Compound Assignment Operators:

A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T) ((E1) op (E2)), where T is the type of E1, except that E1 is evaluated only once.

Thus,

i += lng;

is equivalent to

i = (int)(i + lng);

Why does the Java increment operator allow narrowing operations without explicit cast?

3 Answers3

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