increment operator misbehaving

int ppi = 10,ppj=20,ppk=30,ppl =40;

Code:

---

        printf("\n\nOriginally ppi = %d",ppi);
        printf("\nPassed ppi = %d , and ppi++ = %d", ppi, ppi++);
        printf("\nFinally ppi = %d",ppi);

        printf("\n\nOriginally ppj = %d",ppj);
        printf("\nPassed ppj = %d , and ++ppj = %d", ppj, ++ppj);
        printf("\nFinally ppj = %d",ppj);

        printf("\n\nOriginally ppk = %d",ppk);
        printf("\n++ppk = %d , ppk = %d", ++ppk, ppk);
        printf("\nFinally ppk = %d",ppk);

        printf("\n\nOriginally ppl = %d",ppl);
        printf("\nppl++ = %d , ppl = %d", ppl++, ppl);
        printf("\nFinally ppl = %d",ppl);

---

Output:

Originally ppi = 10
Passed ppi = 11 , and ppi++ = 10
Finally ppi = 11

Originally ppj = 20
Passed ppj = 21 , and ++ppj = 21
Finally ppj = 21

Originally ppk = 30
++ppk = 31 , ppk = 31
Finally ppk = 31

Originally ppl = 40
ppl++ = 40 , ppl = 41
Finally ppl = 41

ppi and ppj appear to be behaving to the proper function passing conventions, but ppk and ppl appear to be misbehaving.. Can anyone pls help me!!!

The C standard doesn't specify which order function arguments are evaluated - it's completely implementation-defined.

Using ++ more than once in any statement is undefined behavior. You must use it a maximum of once.

Quote:

---

Originally Posted by Elysia

Using ++ more than once in any statement is undefined behavior. You must use it a maximum of once.

---

That is not true. For example, the operands could be entirely unrelated variables, or there could be a sequence point between the usage in the same statement.

Ah, bad phrasing.
Modifying the same variable twice or more within a sequence point (statement?) is undefined behavior, it is, I believe?

Thanks guys, for pitching in..
I dont think, it can be completely random though..
It has to follow a pattern, right to left or left to right..
Above appears that compiler is generating a random code?? How can it be.. Am I missing something??

It is random. The compiler can choose its own way of handling it, so there is no "standard" way of doing it.

Thanks guys.. I will try to find out more abt it.. thanks a lot :)

Quote:

---

Originally Posted by Elysia

Modifying the same variable twice or more within a sequence point (statement?) is undefined behavior, it is, I believe?

---

Yes, though in this case I think the second part of that rule, that "the prior value shall be read only to determine the value to be stored", along with the rule that the order of evaluation of arguments is unspecified, comes into play. The former would result in undefined behaviour, so I suppose that the latter is just a matter of "academic interest".