Possible Duplicate:
Why is ++i considered an l-value, but i++ is not?
In C++ (and also in C), if I write:
++x--
++(x--)
i get the error: lvalue required as increment operand
However (++x)-- compiles. I am confused.
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10Please don't do this it hurts my eyes. :) – InsertNickHere Jul 19 '10 at 13:56
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4Possible duplicate of http://stackoverflow.com/questions/371503/why-is-i-considered-an-l-value-but-i-is-not – Péter Török Jul 19 '10 at 13:57
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`gcc` will not compile `(++x)--` here, `g++` will. Same holds for `++i = 5;` – mvds Jul 19 '10 at 13:58
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That really shouldn't compile. What compiler are you using, so I know to avoid it? – nmichaels Jul 19 '10 at 14:05
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In C++ `(++x)--` is syntactically correct but invokes Undefined Behaviour. – Prasoon Saurav Jul 19 '10 at 14:09
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Nathon: yes, it should compile. (The standard definition of operator ++ in a custom iterator class is `iterator & operator++ ()`) – Ken Bloom Jul 19 '10 at 14:09
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@Prasoon: prove to me that's undefined. The semantics of how that works on a custom iterator class are quite clear. – Ken Bloom Jul 19 '10 at 14:10
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@Ken: The variable `x` is being modified more than once between two sequence points and so the behavior is undefined in this case. Correct me if I am wrong. – Prasoon Saurav Jul 19 '10 at 14:12
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1@Prasoon: There's a sequence point before the function call of the `operator--`. There is also a sequence point after the `operator++` returns. – Ken Bloom Jul 19 '10 at 14:25
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3@Ken: Where did he mention that `x` is an object and not a primitive type? `C` the tags :P – Prasoon Saurav Jul 19 '10 at 14:29
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@Ken. Who cares if for some bazare instance it is valid. In most instances it is undefined. But most importantly it is hard to read and understand and for that reason alone it should not be used. – Martin York Jul 19 '10 at 15:34
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This actually raises an interesting question of why we don't define `operator++` to return a const reference. – Ken Bloom Jul 19 '10 at 16:01
1 Answers
10
Post- and pre-increment operators only work on lvalues.
When you call ++i the value of i is incremented and then i is returned. In C++ the return value is the variable and is an lvalue.
When you call i++ (or i--) the return value is the value of i before it was incremented. This is a copy of the old value and doesn't correspond to the variable i so it cannot be used as an lvalue.
Anyway don't do this, even if it compiles.
Mark Byers
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