Inline const expressions have been stabilised (github.com)

submitted 2 months ago by [email protected] to c/[email protected]

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[-] [email protected] 6 points 2 months ago

This is a nice small feature. I'm curious about the commit description:

foo(const { 1 + 1 })

which is roughly desugared into

struct Foo;
impl Foo {
    const FOO: i32 = 1 + 1;
}
foo(Foo::FOO)

I would have expected it to desugar to something like:

foo({
  const TMP: i32 = 1 + 1;
  TMP
})

But I can't seem an explanation why the struct with impl is used. I wonder if it has something to do with propagating generics.

[-] [email protected] 10 points 2 months ago

It's because it has to work in pattern contexts as well, which are not expressions.

[-] [email protected] 2 points 2 months ago

Wait, in pattern context? How? Can you give an example?

[-] [email protected] 10 points 2 months ago

fn foo(x: i32) {
    match x {
        const { 3.pow(3) } => println!("three cubed"),
        _ => {}
    }
}

But it looks like inline_const_pat is still unstable, only inline_const in expression position is now stabilized.

[-] [email protected] 7 points 2 months ago

They tested the same strings on that implementation

The code they were looking at was used for writing the table, but they were testing the one that read it (which is instead correct).

though judging by the recent comments someone’s found something.

Yeah that's me :)The translation using an associated const also works when the const block uses generic parameters. For example:

fn require_zst<T>() {
    const { assert!(std::mem::size_of::<T>() == 0) }
}

This can be written as:

fn require_zst<T>() {
    struct Foo<T>(PhantomData<T>);
    impl<T> Foo<T> {
        const FOO: () = assert!(std::mem::size_of::<T>() == 0);
    }
    Foo::<T>::FOO
}

However it cannot be written as:

fn require_zst<T>() {
    const FOO: () = assert!(std::mem::size_of::<T>() == 0);
    FOO
}

Because const FOO: () is an item, thus it is only lexically scoped (i.e. visible) inside require_zst, but does not inherit its generics (thus it cannot use T).

[-] [email protected] 5 points 2 months ago

Huh, this is awesome! From the link, this now works:

let v: Vec<i32> = const { Vec::new() };

I'm going to have to play with this to see how far it goes.

[-] [email protected] 2 points 2 months ago* (last edited 2 months ago)

What I'm curious about - this function was already const, so for stuff like this, I'd think there's basically a 100% chance the compiler would optimize this too, just implicitly.

AFAIK this new feature is just for times when it isn't an optimization, but more your own domain invariants. E.g. assertions.

But I could be wrong. I wonder if this can be used for actual optimizations in some places that the compiler couldn't figure out by itself.

[-] [email protected] 2 points 2 months ago

Wouldn't this just prevent you from allocating more memory (than zero)?

[-] [email protected] 2 points 2 months ago* (last edited 2 months ago)

Ah, apparently for now you're not allowed to allocate. But vec::new_in(allocator) looks interesting. This works in nightly today:

#![feature(allocator_api)]

use std::alloc::Global;

fn main() {
    const MY_VEC: Vec<i32> = const {
        Vec::new_in(Global)
    };
    println!("{:?}", MY_VEC);
}

Maybe at some point I can append to it at compile time too. I'd love to be able to put a const {} and have allocations that resolve down to a 'static, and this seems to be a step toward that.

I guess I'm just excited that Vec::new() is the example they picked, since the next obvious question is, "can I push?"

[-] [email protected] 0 points 2 months ago* (last edited 2 months ago)

I'm not getting it. What's the point? It seems very much like a cpp-ism where you can put const in so many places.

const int n2 = 0;    // const object
int const n3 = 0;    // const object (same as n2)
// https://learn.microsoft.com/en-us/cpp/cpp/const-and-volatile-pointers?view=msvc-170
const char *cpch;  // const variable cannot point to another pointer
char * const pchc; // value of pointer is constant

int f() const; // members cannot be modified in this, only read
std::string const f(); // returns a constant

Then there are constant expressions.

Can anybody look at that and tell me it's readable with a straight face? I hope they don't start adding all this stuff to rust.

Anti Commercial-AI license

[-] [email protected] 4 points 2 months ago

It can be used for producing const values in arbitrary context. Can basically be swapped for c++'s constexpr.

C++'s const does not exist in rust (values are const by default).

[-] [email protected] 3 points 2 months ago

Nope. This little neat feature mainly is just necessary part of bigger one - const-generics with const bounds.

this post was submitted on 25 Apr 2024

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