“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 accessed only to determine the value to be stored.”

There is a sequence point before evaluating the arguments to a function, and a sequence point after all the arguments have been evaluated (but the function not yet called). Between those two (i.e., while the arguments are being evaluated) there is not a sequence point (unless an argument is an expression includes one internally, such as using the && || or , operator).

That means the call to printf is reading the prior value both to determine the value being stored (i.e., the a++) and to determine the value of the second argument (i.e., the a++) and also 3rd argument (i.e the ++a). This clearly violates the requirement quoted above, resulting in undefined behavior.

compiler will give warning as its undefined behaviour. for operation on a.

output of this program will differ on different system ..!!