EDIT: Okay, got it.
The first point to make is that obviously you shouldn’t use this code anyway. However, when you expand it, it becomes equivalent to:
j = j ^ (i = i ^ (j = j ^ i));
(If we were using a more complicated expression such as foo.bar++ ^= i, it would be important that the ++ was only evaluated once, but here I believe it’s simpler.)
Now, the order of evaluation of the operands is always left to right, so to start with we get:
j = 36 ^ (i = i ^ (j = j ^ i));
This (above) is the most important step. We’ve ended up with 36 as the LHS for the XOR operation which is executed last. The LHS is not “the value of j after the RHS has been evaluated”.
The evaluation of the RHS of the ^ involves the “one level nested” expression, so it becomes:
j = 36 ^ (i = 25 ^ (j = j ^ i));
Then looking at the deepest level of nesting, we can substitute both i and j:
j = 36 ^ (i = 25 ^ (j = 25 ^ 36));
… which becomes
j = 36 ^ (i = 25 ^ (j = 61));
The assignment to j in the RHS occurs first, but the result is then overwritten at the end anyway, so we can ignore that – there are no further evaluations of j before the final assignment:
j = 36 ^ (i = 25 ^ 61);
This is now equivalent to:
i = 25 ^ 61;
j = 36 ^ (i = 25 ^ 61);
Or:
i = 36;
j = 36 ^ 36;
Which becomes:
i = 36;
j = 0;
I think that’s all correct, and it gets to the right answer… apologies to Eric Lippert if some of the details about evaluation order are slightly off 🙁