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i is increased multiple times before the next sequence point, so it is undefined behavior. On a different platform /compiler the result could be totally different

Well this statement is the violation of point rule and hence the behavior is undefined.

Different compilation provide different rules and the output cannot be determined to the accuracy.

me running print value is 322

The right is 3 2 2 i think, the expression is evaluated right to left.
so ++i becomes 2
I++ remains 2
i will be 3.
Then the print statement will print values as 3 2 2.

Forgot to mention, printf statements like above are like stack operations so when the compiler sees this expression, it reads each value and pushes into the stack (push operation). When it needs to do the evaluation, it pops out each of the values from the stack,hence the right to left evaluation.

recoil.rahul,
What's your platform?
And don't you all think it has got nothing to do with the precedence of i++ over ++i ?