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.
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.)
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.)
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.
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:
> 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.
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.)
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:
K0IN | 10 hours ago
sheafification | 9 hours ago
https://en.cppreference.com/cpp/language/reinterpret_cast
EDIT: If you’re able to use C++20 then clearly bit_cast is better.
jasode | 9 hours ago
https://stackoverflow.com/questions/53401654/why-was-stdbit-...
leni536 | 9 hours ago
antiloper | 9 hours ago
What's so genuine about this?
tialaramex | 9 hours ago
Because neither of these are memory safe languages, you are required to ensure you've made no mistakes or else anything might happen.
ndr | 9 hours ago
> and most blog posts on the topic get it wrong.
This smells like Claudism/LLMish.
eqvinox | 9 hours ago
(Feels a bit like LLM junk, but honestly not sure.)
LoganDark | 9 hours ago
Feels like LLM to me. I looked at some of the rest of the article and... pretty much certain now.
rfgplk | 8 hours ago
Why leave such disparaging comments?
munificent | 8 hours ago
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 | 8 hours ago
pipe01 | 9 hours ago
scoopr | 9 hours ago
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 | 8 hours ago
leni536 | 8 hours ago
Also it seems that bit_cast in particular is going to be strengthened in this regard:
https://cplusplus.github.io/LWG/issue4539
StilesCrisis | 8 hours ago
I can't downvote posts yet, but if you have the ability, consider it. This is clearly LLM slop without human review.
eqvinox | 8 hours ago
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 | 7 hours ago
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 | 8 hours ago
Shouldn't the code return 3 in the base case and 4 if the check for 2 was assumed to always hold ?
bena | 8 hours ago
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 | 6 hours ago
Martin_Silenus | 7 hours ago