eqvinox 12 hours ago

The example included below "Why C and C++ Differ" is UB in both C and C++ (and says nothing about "why" C & C++ differ). The article isn't overall wrong but that bit is just...

(Feels a bit like LLM junk, but honestly not sure.)

  • LoganDark 12 hours ago

    > A pointer cast worked fine at -O0 and silently broke at -O2.

    Feels like LLM to me. I looked at some of the rest of the article and... pretty much certain now.

  • rfgplk 11 hours ago

    > (Feels a bit like LLM junk, but honestly not sure.)

    Why leave such disparaging comments?

    • munificent 11 hours ago

      Because if the comment is correct, it's useful signal for other potential readers to know that they can skip it.

      I try to be as charitable as possible when reading and commenting online, but we don't have infinite attention and thanks to AI, it's easier than ever to end up wasting it on things that aren't worth reading.

      (I'm not claiming that the article here is or is not an example of that.)

    • quietbritishjim 11 hours ago

      Even if it's not from an LLM, it's still wrong! You've focused on the last important of that comment, really.

scoopr 12 hours ago

I just watched video[0] that argues if you can consteval something, then it must be free of UB, which seems compelling, at least for these lower level helpers.

Unfortunately, it seems memcpy is not constexpr, so cannot be used like this, but std::bit_cast actually works[1] as constexpr, so I think in C++ that would now be the most preferred use. It won't allow the union either.

[0] https://www.youtube.com/watch?v=-LAXqqqX274 [1] https://godbolt.org/z/xGzjTMGvv

  • rfgplk 11 hours ago

    __builtin_memcpy is constexpr under clang with constraints

  • leni536 11 hours ago

    Careful, only language UB is guaranteed to be detected in constant evaluation, UB in standard library functions isn't. It is a good sanity check though.

    Also it seems that bit_cast in particular is going to be strengthened in this regard:

    https://cplusplus.github.io/LWG/issue4539

antiloper 12 hours ago

> The C vs C++ distinction here is genuinely treacherous

What's so genuine about this?

  • tialaramex 12 hours ago

    Despite the fact they're two different languages, people are often taught as though it's basically just a subset relationship and then they carry this through into the actual software they write.

    Because neither of these are memory safe languages, you are required to ensure you've made no mistakes or else anything might happen.

  • ndr 12 hours ago

    It continues with

    > and most blog posts on the topic get it wrong.

    This smells like Claudism/LLMish.

StilesCrisis 11 hours ago

The conclusion is simply wrong. `union` should not be used for type-punning in C++, or in C code which might be compiled by a C++ compiler.

I can't downvote posts yet, but if you have the ability, consider it. This is clearly LLM slop without human review.

  • eqvinox 11 hours ago

    > C code which might be compiled by a C++ compiler.

    I really hope you're talking mostly about header files; for actual code this is such a horrible idea (and in a lot of cases just won't work without massive efforts.) Even for header files, arguably one ought to really know what they're doing.

    • StilesCrisis 11 hours ago

      For sure, a union which is used as API surface to a C library which a C++ codebase uses is the most obvious and dangerous footgun here.

      The article never even discusses how C++ actually implements unions to begin with, so it's just incomplete. (Only the "active" member is meaningful, the others are considered as meaningless and shouldn't be touched.)

LelouBil 11 hours ago

I don't understand the example in "Why C and C++ Differ".

Shouldn't the code return 3 in the base case and 4 if the check for 2 was assumed to always hold ?

  • bena 11 hours ago

    You're halfway there. It's a little more messed up than that.

    IIRC, this is all little-endian. So assume our 8 bytes labeled A-H. For a 32 bit/4 byte integer, they will be read as DCBA. For a 64 bit/8 byte integer, HGFEDCBA.

    So the struct is set up like a = DBCA, b = HGFE. So when we assign 2 and 3 to a and b, it should look like this in memory:

    00000010 00000000 00000000 00000000 and 00000011 00000000 00000000 00000000

    When we cast that to a uint64 and assign 4 to it, we should wind up with:

    00000100 00000000 00000000 00000000 00000000 00000000 00000000 00000000

    Which effectively zeros out b.

    So if the conditional is evaluated, it will evaluate to false, we return b, which is 0.

    If the conditional is not evaluated, we should return the value in a, which is 4.

    • LelouBil 9 hours ago

      Oooh thanks, I missed that the cast was into a 64 bit integer

pipe01 12 hours ago

Is it necessary to use the lowest possible font weight?

K0IN 13 hours ago

Does reinterpret_cast works in c++ for this?

Martin_Silenus 11 hours ago

Hackers don't make Quake III Arena by trusting C rules or compiler optimizations anyway.