Confused with pre and post increment operator

Question

hello I am learning basics of C programming language, recently i have studied about post and pre increment/decrement operators and also about lvalue and rvalue, the following program shows an error, lvalue required, according to me it should give a value of 6, Can anyone please explain why?

int main(){
  int x = 8, y;
  y = --x--;
  printf("y=%d",y);
  return 0;
}

Please explain, why is it so?

kindly mention your comments if you vote down this, this seems to be very tricky for me. — Mayank Tiwari, Jul 03 '13 at 05:11
All it gives you is Undefined behavior. `x` is modified more than once without a intervening sequence point. — Alok Save, Jul 03 '13 at 05:13
`y = --x--` should return `6`? By that logic, `y = --------x--------;` should return `0`, right? — devnull, Jul 03 '13 at 05:14
@devnull, hope so, i am beginner so i am not sure about what you have metntioned in your comments..... — Mayank Tiwari, Jul 03 '13 at 05:15
Actually in your expression, `i` is modified twice without an intervening sequence point, which is undefined behavior in C and C++. This is what @AlokSave commented — Grijesh Chauhan, Jul 03 '13 at 05:36
@OliCharlesworth, check it once more, it is not a duplicate question at all, check the error which this program is having, u will again feel that it is not a duplicate question.... i hope it is clear to you........ — Mayank Tiwari, Jul 03 '13 at 06:13

Pavel Dudka · Accepted Answer · 2013-07-03T05:23:22.133

6

Well, let's see what is happening in --x--. At first, post-decrement executes: --(x--). (x--) = 7. After that result of this operation is placed to the original structure: --7 - doesn't make sense - thus you get lvalue required error

edited Jul 03 '13 at 05:23

answered Jul 03 '13 at 05:16

Pavel Dudka

20,754
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1

Doesn't postfix have precedence over prefix? – llakais Jul 03 '13 at 05:21
1

You'r right. forgot about that :(. Will update an answer – Pavel Dudka Jul 03 '13 at 05:22
Haha cool. The only reason I mentioned it is in C++ (I know this question is tagged C), I think `(++x)++` would actually work, since in C++ prefix operators return lvalues. – llakais Jul 03 '13 at 05:26

Deepu · Answer 2 · 2013-07-03T05:26:22.627

5

The statement y = --x--; will give you the following error on compilation in C. lvalue required. This is because the post decrement operator -- will return an rvalue after operating on the variable x. So there are no lvalue to perform the pre decrement operator -- afterwards.

But this is one point where C and C++ differs. In C the following statement will also give you the same error lvalue required.

y = (--x)--;

But in C++ the statement y = (--x)--; will compile fine and the value of y is 7. Because unlike C, C++ returns an lvalue after performing the pre decrement operator on variable x.

edited Jul 03 '13 at 05:26

answered Jul 03 '13 at 05:17

Deepu

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+1 And, GCC says: `error: expression is not assignable` with an underlined `x--`. – meaning-matters Jul 03 '13 at 05:18

Dayal rai · Answer 3 · 2013-07-03T05:26:49.227

L Value is left operand of assignment operator which should refer to memory location.As explained by @Pavel your Lvalue is becoming a value not object so you are getting error.

--x means x = x-1 but in your case it is becoming --7 which will be equivalent to 7 =7-1 which is definitely not valid expression.

Other than this more than one operation on same variable without any sequence point in between, results undefined behaviour.

score 1 · Answer 4 · answered Jul 03 '13 at 05:20

C order of operations specifies that postfix operators have precedence over the prefix operators. The -- postfix operator returns the current value (an rvalue) of the operand and then decrements the operand. Then the prefix decrement operator would be applied...but the decrement/increment operators need lvalue operands since they by definition modify their operands. So, as the compiler says, an lvalue is required.

score 1 · Answer 5 · answered Jul 03 '13 at 06:23

You should not use it at a time because you will not understand the behavior of compiler. So, you need to guide your code so that they will forced to do what you like.

Now come to your point. If you want to decrease the value by one you can use a-- or --a. They will do the same. If a = 5 and you use b=a-- you will get b = 5 and a = 4 where if you use b=--a you will get b = 4 and a = 4 in --a the value is assigned immediately and in a-- value will be assigned after statement is complete. Hope you are clear.

L value required error shown when it doesn't find any suitable variable where it can be assigned.

Confused with pre and post increment operator

5 Answers5