What I mean is, what’s to stop us saying that C upholds all the same guarantees that Rust does and that it’s the programmer that’s responsible for upholding them (just as the programmer is responsible in the case of Rust code marked ‘unsafe’)? This seems like a semantic game to avoid acknowledging that unsafe Rust comes with some of (though not all) of the same risks as C code.

In short, the definitions are not important. What matters are the risks that you do or don’t run. And if your Rust code contains unsafe blocks, you are running risks that you wouldn’t be if you used Fil-C, which has no such escape hatch. (Of course this goes both ways – your Fil-C code is more likely to fail, safely, with a runtime error due to a mistake that Rust would have caught at compile time.)

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tialaramex 4 hours ago | parent [-]

And do you say that C offers these guarantees ?

Real world C software does not read like software written by people who are in fact upholding those guarantees you say C could equally have. It reads as though they think such a guarantee is a joke or an irrelevance. It's not rare for me to run into people who think C's pointers are just indexing into a massive array of all RAM (or its equivalent on today's systems with virtual addressing), that's not just not in the same ballpark as a safe C program, that's playing a different sport on another continent.

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foldr 3 hours ago | parent [-]

You seem to be suggesting that a language being safe or unsafe is a social contract rather than a technical property of the language.

>And do you say that C offers these guarantees ?

No, that would be silly, and it's an illustration of why it is silly to say that a language guarantees X if it is the programmer who must check that X holds. If we go down that route (which, to repeat, would be silly), then we can make C safe without any technical changes just by adding some language to the standard saying that C programmers are obliged to ensure that their code maintains a certain list of invariants. When you say that "Rust makes the same guarantees regardless of the unsafe keyword", it seems to me that you are doing something equally pointless.

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bit1993 an hour ago | parent [-]

> ... then we can make C safe without any technical changes just by adding some language to the standard saying that C programmers are obliged to ensure that their code maintains a certain list of invariants.

In Rust you can use #![forbid(unsafe_code)] to totally forbid unsafe code in your codebase. Rust also checks for memory safety at compile time, these are strong guarantees that ensure that if the code compiles it is memory safe.

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foldr an hour ago | parent [-]

I'm aware of that, but I'm responding to the original claim that "Rust makes the same guarantees regardless of the unsafe keyword" (see https://news.ycombinator.com/item?id=46262774)

	▲	bit1993 an hour ago \| parent [-]
		Ah. I agree with you. When unsafe is used the borrow checker cannot check for memory safety, the programmer has to provide the guarantees by making sure their code does not violate memory safety, similar to programming in C. But unsafe Rust is still far better than C because the unsafe keyword is visible and one can grep it and audit the unsafe parts. Idiomatic Rust also requires that the programmer provides comments as to why that part is unsafe.