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Could anyone explain these undefined behaviors (i = i++ + ++i , i = i++, etc…)
int i = 10;
int n = i++*5*i;
Output
value of n = 550 (in Java) value of n = 500 (in C and C++ )
Why not same result? Why different?
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3Could it be because they're different languages? – R. Martinho Fernandes Sep 16 '11 at 00:11
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13Anyone who attempts to use code like that in any language should be taken out back and shot. – Hovercraft Full Of Eels Sep 16 '11 at 00:12
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@Hovercraft Full Of Eels: not much different to some lambdas I've seen.... – Mitch Wheat Sep 16 '11 at 00:20
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I think people are surprised when a mature, common language such as C or C++ react in an unpredictable way. It's so contrary to the usual programming experience. That's why this question keeps coming up. – Mark Ransom Sep 16 '11 at 16:11
3 Answers
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In Java, this is a well defined operation. It will:
- increment
i(it's now 11); - Produce the old value of
i(10), because you used the postfix increment operator; - Multiply that by 5 (10*5 = 50);
- Multiply that by the current value of
i(50*11 = 550);
In both C and C++ this operation has undefined behaviour, so anything could happen. If anything could happen, that explains the results, whatever they are, and whether they make sense to you or not.
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R. Martinho Fernandes
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In C and C++, operations such as:
j = i++ + i;
are undefined, due to lack of sequence points. In Java, they are well defined. Therefore, you could see a difference in results.
Oliver Charlesworth
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Forgive me for nitpicking, but that's not a correct use of "therefore" - he could easily have seen the same results, they just happened to be evaluated in a different order. In other words, '550' could well be output by the C++ code. – Arafangion Sep 16 '11 at 00:20
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I thought that the operators were handled with same priorities in C C++ and Java but it seems that Java handles differently. In the context of JAVA ++ has more high priority than % as documentation http://w3i.in/754 int i = 10; int n = i++*5*i; in this case i++ is done but if expression is as follows: int i = 10; int n = i++*5; then i++ is handled after multiplication even though multiplication has lower priority. why? – Anand Sep 16 '11 at 00:21
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3@Arafangion: "Therefore" seems fine to me. The first sentence is a statement; the second statement is the conclusion. (Note that I was careful to say "could", not "will".) – Oliver Charlesworth Sep 16 '11 at 00:22
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2@Anand it's not possible to reason about this in terms of operator precedence. You must understand what a [sequence point](http://en.wikipedia.org/wiki/Sequence_point) is. – Karl Knechtel Sep 16 '11 at 00:38
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Because what you are doing is undefined. The increment operator should not be placed in expressions of assignment with the variable being incremented.
i = i++; //undefined
n = i++ + i; // also undefined
zellio
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He's not using it as such. He's doing (i++) * 5 * i. The question is when does i get incremented. – Hovercraft Full Of Eels Sep 16 '11 at 00:11
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2Yes, this answer is nonsense (apart from the part that says that this is undefined). – Oliver Charlesworth Sep 16 '11 at 00:14
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@Benjamin nothing; but that's because `n` is not "the variable being incremented", so it isn't a case that the answerer is describing. – Karl Knechtel Sep 16 '11 at 00:39
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@Karl: My comment was in response to the second version of this answer, you're looking at the third. – Benjamin Lindley Sep 16 '11 at 01:06
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@Karl: but it is a lot more like the case that the *questioner* was describing. The question does no have `i` on both sides of the assignment. – R. Martinho Fernandes Sep 16 '11 at 14:54