For context: I am trying to write a Rust wrapper over a C library.
Like many C libraries, most of its functions return an int
. Positive return values are meaningful (provides information) and negative values are error codes.
To give an example, think of something like int get_items_from_record(const struct record *rec, struct item *items)
. A positive value indicates how many items were returned. -1
could mean ErrorA, -2
ErrorB, and so on.
Since this is Rust, I want to represent this kind of integer as Result<T, E>, e.g.:
enum LibError {
A = -1,
B = -2,
// ....
}
// LibResult is ideally just represented as an integer.
type LibResult = Result<NonNegativeInteger, LibError>;
// Then I can pass LibResult values back to the C code as i32 trivially.
Is there a way/crate to do this?
It matters because the conversion between i32 and the Result is only “free” if they have the same layout (which they do not, because of the discriminant). So a more costly conversion method is required.
You are right, because the compiler is able to optimize your code quite well. However, if that optimization were to break at some point (as there is no guarantee that an optimization will continue to work in the future), it would become less efficient.
That seems like strong premature optimisation. Perhaps worth a note, but I’d presume the majority of people the majority of the time wouldn’t need to worry about that.
For real. Unless he’s converting between results and ints millions of times a second, I think he’s going to be just fine using the idiomatic solution. That transmute shit I’d wack lol
So what! Who cares if it’s free? Write first, profile and optimize later. Not everyone cares about whether the conversion is free. Simply matching and converting to the right integer is fast enough.
The reason I asked the question, was that I wanted to keep an
int
anint
throughout the program.It’s not for performance reasons, it’s just that I feel like there is a certain elegance in keeping things type safe entirely “for free” much like how
Option<&T>
is actually just a regularT*
, even if it could be pointless in the big picture.Well if you want to keep an int as an int, then use an int. If you want to use Results, use results. Like I said to the other commenter: unless this code is in a very tight loop where performance is crucial, you’ll be fine just implementing From/Into to change your Result to and from an int. You’re already crossing a language boundary by calling c code from Rust. Is it that much more of an issue to just convert the types when you need to?
Trying to store the results as an int in some magical way just screams premature optimization. You have abstraction tools at your disposal. They may not be zero cost, but they are cheap. Use them.
First and foremost: It’s not about optimization, as I have mentioned before. Never once have I intended to optimize the conversion, because I know it is pointless. Stop making that assumption and only then we can continue the discussion.
There is no reason why people cannot use Rust as “C, but actually type-safe”. A type-safe representation of C’s error code pattern is a part of that. This way the code is “backwards compatible” with debuggers designed for C/C++, such that “-EINVAL” is actually displayed as “-EINVAL” in the debugger and not a mysterious struct of
(discriminant, data)
.Oh I see. I misunderstood the reason for wanting it represented as an int.
I’m wondering if you could just create a wrapper type that only has an int as a member, but then implement a trait on it so that it can act like a result. That, or just pass around your int type in the rust code, and when you need it to act like a result you do a conversion from int to result. Your debugger wouldn’t show it as an int at that point, but it wouldn’t show any other Result as an int anyway so it would br consistent with other rust code. If this still doesn’t work, you could even make a struct that contains both the int and the result and keeps then synchronized. Then, when debugging, you could look into that struct and see the int value like you want.