Reply to thread

Message: [QUOTE="ena8t8si, post: 2470779"] All ok up to this point. This step is where your reasoning goes wrong. Consider this sequence of evaluation, writing {f(1)} to mean the temporary value that resulted from a previous evaluation of f(1), and similarly {f(0)}, {f(2)}, {f(3)}: f(0) (*{f(0)})++ f(2) (*{f(2)})++ f(1) f(3) (*{f(1)})++ (*{f(3)})++ For the last two lines, {f(1)} == {f(3)}. There are no sequence points, but both lines update what the pointers point to, in other words they update the same object. That's undefined behavior. The relevant side effect operators are outside the function calls, and both side effects may be done after both function calls have completed. It's the same, as I have explained above. [/QUOTE]