how does increment work?

Question

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Could anyone explain these undefined behaviors (i = i++ + ++i , i = i++, etc…)
Undefined Behavior and Sequence Points

Ok we all know that i++ increments value by 1 on next line and ++i increments on same line
(please correct me if i am wrong there )
so for a sample statement of c as follows:

int a=0;  
printf("%d , %d",++a,a);  

the expected output should be 1 , 1 but instead it gives 1 , 0 so as one might guess what i am asking here is why does the second linking of i print 0 instead of 1 when the value is already incremented.

So if post increment didn't increment value in the same line then what is the difference between post and pre increment?
edit : changed the name of variable from i to a to avoid grammatical confusion.

Answer 1

It's undefined behavior. the compiler is allowed to calculate the parameters in any order. your compiler just calculate it from right to left, so the rightest parameter is 0, and the 2nd is 1.

edit: as Seth said, the compiler is only free to change the order of calculating, not to do whatever it wants, so when you don't care about the order you can freely call functions, but you should never assume that one parameter is been calculated before another.

Answer 2

I think your question is not about how increment work. The phenomenon you have observed is about the order of passing parameter to a function. As far as I can remember, this is not defined by c++ standard, so it's a UNDEFINED BEHAVIOR. Meaning different compiler could have different implementation. E.g. One compiler could pass parameter from left to right, then a different could pass parameters from right to left.

You can have a better read of Sequence point here: http://en.wikipedia.org/wiki/Sequence_point

Answer 3

printf("%d , %d",++i,i);

is undefined behavior.

Your are violating C sequence points rule:

(C99, 6.5p2) "Between the previous and next sequence point an object shall have its stored value modified at most once by the evaluation of an expression. Furthermore, the prior value shall be read only to determine the value to be stored."

Answer 4

You are very wrong in your understanding of the increment operators. i++ increments i, and returns its old value; ++i increments i, and returns the new value. When the actual incrementation takes place is only guaranteed to be after the preceding sequence point, and before the next; in your code, this means before the call to printf.

Beyond that (and largely because of that), if you modify the value of an object, you're not allowed to access it anywhere else without an intervening sequence point, except as needed to determine the new value. You violate this rule, so your code has undefined behavior.

Answer 5

Once I read in some book that when a statement is like this:

printf("%d , %d",++a,a);

The execution starts from right to left but outputted as left to right. Therefore you see the out as 1 0 - 'a' is calculated first = 0 then '++a' = 1, but displayed as 1 0

This is strange but yeah that's how it works.

Hope this helps.