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by jcr1488 2604 days ago
Right, but I didn't make a case that memset() is any more secure, did I? The parent comment was talking about undefined behaviour as if it's some kind of universal get-out clause.
1 comments

Your last paragraph strongly implies that hermitdev was using undefined behavior to "explain away" security concerns.

But the problem is that the security concerns and fixes are all undefined here.

So the initial comment is still very right. If you're relying on the padding to be anything in particular, you're in trouble.

> But the problem is that the security concerns and fixes are all undefined here.

No one in this sub-thread has mentioned (or implied) any "fix". You appear to be putting words in my mouth.

> So the initial comment is still very right.

Of course it's right, but it's also misleading when read in the context of the parent comment.

> If you're relying on the padding to be anything in particular, you're in trouble.

No shit.

> No one in this sub-thread has mentioned (or implied) any "fix". You appear to be putting words in my mouth.

I'm talking about the comments hermitdev was replying to that were treating memset as a 'fix'.

And 'fix' is shorthand for the opposite of "open[ing] you up to all kinds of info leaks". I don't think that's putting words in anyone's mouth.

I guess we are misunderstanding each other's point. I just found hermitdev's comment to be misleading (despite being correct), but perhaps it's just my reading of it.

To be fair though, memset() usually IS a fix. As mentioned by the kernel memzero_explicit() docs:

> usually using memset is just fine (!)

-- https://www.kernel.org/doc/htmldocs/kernel-api/API-memzero-e...

A conforming C compiler can't just remove memset() as it pleases. The case that most often requires memzero_explicit() is when zeroing an object after destruction, because the compiler thinks it can statically determine that it's a dead store. It very rarely happens that a compiler elides a memset() used for initialization.

I'm not sure why you seem to think that memset() can just be dropped at will for no reason whatsoever or that it's somehow always undefined behaviour.

A conforming C compiler can remove a memset that has no side effects.
You mean it can remove a memset() that doesn't cause the observable behaviour to change?

For the sake of argument, can you show me some example code where it would be conforming to remove a memset() call? Preferably a realistic example and not a Google'd copypasta. Because it's all too easy to just regurgitate things you heard and think you understood, but no so easy to demonstrate it yourself.