Can anyone explain in depth why this outputs 9
here is my code in PHP:
$x = 4;
$x = $x+++$x++;
echo $x;
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2why are you insisting this should be tagged with `java` and `c`? – Weather Vane Nov 26 '15 at 18:15
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because its about order of execution not about language. – Aadil Ali Nov 26 '15 at 18:17
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1It is language-specific! Do not add tags for unrelated languages. – too honest for this site Nov 26 '15 at 18:18
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@Olaf Question is equally applicable to all 3 languages, the syntax is just slightly different, i.e. Java and C doesn't use `$`, is all. – Andreas Nov 26 '15 at 18:20
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I added $ to describe x is a variable. – Aadil Ali Nov 26 '15 at 18:22
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That would not make any sense in `C` or `Java`. You should've tagged it as `php` and only `php`. – Emz Nov 26 '15 at 18:23
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1@Andreas: So it is undefined behaviour in all three languages? Then your answer is wrong and has to be downvoted. – too honest for this site Nov 26 '15 at 18:23
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1It is undefined behaviour in C. http://programmers.stackexchange.com/questions/153386/why-is-x-x-undefined – Weather Vane Nov 26 '15 at 18:24
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@Emz So you are saying that there is no increment operator in C or java :D :D – Aadil Ali Nov 26 '15 at 18:26
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`$x+++` there is no such thing in `C` or `Java` no. – Emz Nov 26 '15 at 18:27
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increment operator is same in java x++ + x++ gives same output in java – Aadil Ali Nov 26 '15 at 18:33
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@AadilAli compiled as C code with MSVC, it prints `10`. You can "explain" the result of Undefined Behaviour, but it is still that. – Weather Vane Nov 26 '15 at 18:37
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@Olaf Sorry, was unaware that it's undefined in C. It is however very well-defined in Java. – Andreas Nov 26 '15 at 18:48
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@Emz `x = x+++x++;` is valid in Java, and behavior is well-defined. – Andreas Nov 26 '15 at 18:53
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@Andreas: Just be careful expecting the same semantics from the same syntax. – too honest for this site Nov 26 '15 at 18:56
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@Andreas, again, I never said it wasn't. I said that `$x+++` wasn't in either C or Java, as far as I know it is only in php. – Emz Nov 26 '15 at 18:56
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What is your best guess? – Toby Allen Nov 26 '15 at 19:47
1 Answers
4
Execution goes like this:
$x = $x++ + $x++; ($x = 4)
$x = 4 + $x++; ($x = 5)
$x = 4 + 5; ($x = 6)
$x = 9;
For a more detailed answer of a more complex example in Java, see this answer: Incrementor logic
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The variable is same why not the value change from 4 to 5 before addition ? – Aadil Ali Nov 26 '15 at 18:11
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Value of `x` is actually `6` *right before* the addition, as I've illustrated. – Andreas Nov 26 '15 at 18:12
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