[SAI_Peregrinus]

You can write a C compiler with exactly laid out well-defined semantics. You can't assume those semantics hold for C-the-language, because it doesn't define those semantics. UB is a property of the language, not just of a given compiler. The Rust reference defines the semantics of the safe subset of Rust without any UB, so any compliant Rust compiler won't have UB in that subset. The reference also defines the guarantees which the programmer must uphold within `unsafe` blocks to avoid UB, as long as those are upheld there's no UB at all.

[jstimpfle]

I understand that. It makes no practical difference. 99,99% of my additions don't rely on signed overflow for example, and if I'd ever need it there are ways to get just it.

Or tell me how you write a Rust program differently given that signed overflow is apparently defined? I bet you write it exactly the same way, and you get pretty much the same behaviour in practice. And we're even only debating actual overflow situations, meaning there is a bug whatever the compiled behaviour is.

C the language doesn't even guarantee that the machine has native integers with 8, 16, 32, 64 bits etc, that a cacheline is 64 bits, that a page is 4K, and here I am, writing programs for exactly that.