Is reading an unsigned char array through an unsigned integer undefined and thus not safe?

Question

I read this question and got extremely confused on what I have to do with my code. If my reading is correct, my code below is undefined according to the current standard. Below is everything I need, and I really cannot find a better way to do this. Using left shift with OR is definitely not what I want. Is it just fine to leave it this way?

EDIT: And yes I am reading n after setting the values of rbg and not n.

union Pixel {
    unsigned char rgb[4];
    uint32_t n;

    Pixel():
    n(0) {}
};

Nothing wrong this far. But if you access a member you didn't put there it is UB. — nwp, May 28 '15 at 19:18
@LightnessRacesinOrbit I want to be sure that my code does not contain undefined behaviour. — , May 28 '15 at 19:26
@xiver77: When? Under what conditions? You don't show how you use this `union`. — Lightness Races in Orbit, May 28 '15 at 19:30
Although it is technically UB, many compilers support it as an extension, mainly because - as far as I know - this kind of type punning is allowed in C. An often suggested alternative is to use `memcpy`. — dyp, May 28 '15 at 19:32
http://stackoverflow.com/questions/15952204/using-char-array-inside-union#comment22733039_15952308 — Lightness Races in Orbit, May 28 '15 at 19:33
@xiver77: I still see no usage code. Never mind; the above link answers your question. — Lightness Races in Orbit, May 28 '15 at 19:33
Can you identify your reason that masking and shifting is "...definitely not what I want..."? — 2785528, May 28 '15 at 21:47

score 1 · Accepted Answer · answered May 28 '15 at 20:54

This is one of the parts of C++ where it is easier to check what your compiler is doing than following the standard. As far as I know most compilers still do what you expect (MSVC, GCC, CLANG). Are you thinking something esoteric?

In the following link the Sun C++ compiler is mentioned as not working:

http://blog.regehr.org/archives/959

Is reading an unsigned char array through an unsigned integer undefined and thus not safe?

1 Answers1