Which of the following combinations of post & pre-increment operators have undefined behaviour in C?

Question

I've read, Could anyone explain these undefined behaviors (i = i++ + ++i , i = i++, etc...) and tried understanding Sequence points on "comp.lang.c FAQ" after wasting more than 2 hours of time trying to explain the following results by gcc compiler.

expression(i=1;j=2)     i       j       k
k = i++ + j++;          2       3       3
k = i++ + ++j;          2       3       4
k = ++i + j++;          2       3       4
k = ++i + ++j;          2       3       5

k = i++ + i++;          3               2
k = i++ + ++i;          3               4
k = ++i + i++;          3               4
k = ++i + ++i;          3               6

i = i++ + j++;          4       3
i = i++ + ++j;          5       3
i = ++i + j++;          4       3
i = ++i + ++j;          5       3

i = i++ + i++;          4
i = i++ + ++i;          5
i = ++i + i++;          5
i = ++i + ++i;          6

Question:

1. I want to know if all the expressions shown (in 4 groups) in above figure have undefined behavior? If only some of them have undefined behavior which ones does and which ones doesn't?

2. For defined behaviour expressions, kindly can you show (not explain) how compiler evaluates them. Just to make sure, if I got this pre-increment & post increment correctly.

Background:

Today, I've attended a campus interview, in which I was asked to explain the results of i++ + ++i for a given value of i. After compiling that expression in gcc, I realized that the answer I gave in interview was wrong. I decided not to make such mistake in future and hence, tried to compile all possible combinations of pre and post increment operators and compile them in gcc and then try to explain the results. I struggled for more than 2 hours. I couldn't find single behaviour of evaluation of these expressions. So, I gave up and turned to stackoverflow. After little bit of reading archives, found that there is something like sequence point and undefined behaviour.

Answer 1

Except the first group, all expressions in the other three groups have undefined behaviour.

How the defined behviour is evaluated (group 1):

i=1, j=2;

k=i++ + j++; // 1 + 2 = 3
k=i++ + ++j; // 1 + 3 = 4
k=++i + ++j; // 2 + 3 = 5
k=++i + j++; // 2 + 2 = 4

It's fairly straight forward. post-increment vs pre-increment thing.

In group 2 and group 4, it's quite easy to see the undefined behaviours.

Group 2 has undefined behaviour because = operator doesn't introduce a sequence point.

Answer 2

There are no sequence points within any of these statements. There are sequence points between them.

If you modify the same object twice between consecutive sequence points (in this case, either via = or via prefix or postfix ++), the behavior is undefined. So the behavior of the first group of 4 statements is well defined; the behavior of the others is undefined.

If the behavior is defined, then i++ yields the previous value of i, and as a side effect modifies i by adding 1 to it. ++i modifies i by adding 1 to it, and then yields the modified value.