Does mutably borrowing a const struct completely clone it?

1

I was quite surprised that this code compiles:

#[derive(Debug)]
struct X {
    a: i32,
    b: i32,
}

const SOME_CONST: [X; 2]=[
    X {
        a: 0,
        b: 0,
    },
    X {
        a: 0,
        b: 0,
    },
];

fn function(v: &mut [X; 2]){
    v[0].a = 10;
    println!("{:?}", v);
}

fn main() {
    function(&mut SOME_CONST);
}

How can this work? Is SOME_CONST inlined and a completely new struct created? I would have expected that I need to call clone() because this sounds expensive.

I stumbled upon this in Making Safe Things From Unsafe Parts - Cliffle in the section "Removing the unsafe block from main". Changing a static mut to such a const + local variable makes the program faster, which I do not understand, either. I guess that now the struct is on the stack instead of some static RAM section, but why does this make it faster? In the section " Getting the static out" this made the program slower because Making Safe Things From Unsafe Parts - Cliffle.

2

Basically yes, constants are created anew each time they are mentioned. Mutations to that temporary value will not be seen in other instances of that same constant.

3

No, clone it not called. const works just like C #define with some extra type information. You main function will literally turn into

fn main() {
    function(&mut [
    X {
        a: 0,
        b: 0,
    },
    X {
        a: 0,
        b: 0,
    },
]);
}
1 Like
4

It literally involves (at most) moving around 4 integers. That's not expensive by any standards today, even on low-resource systems.

5

Of course it's not expensive in my example, but with bigger structs.

6

In the case of const values, the copy is done "at compile time" :slight_smile:

2 Likes
7

If your values are so big that a memcpy is a performance problem, you shouldn't be declaring them as const. Consider using static or lazy_static! instead.

8

For big constants, I'd recommend defining it as a reference, then the copied instances are just pointers to the same static memory.

2 Likes
9

Like this:
const SOME_CONST: &[X; 2]=&[...
?

But the complete struct has to have been loaded into RAM (because I could mutate it). How can this be done at compile time?

10

Yes, exactly.

11

Your code gets transformed into (something like) this at compile time:

#[derive(Debug)]
struct X {
    a: i32,
    b: i32,
}

fn function(v: &mut [X; 2]){
    v[0].a = 10;
    println!("{:?}", v);
}

fn main() {
    let mut x = [
        X {
            a: 0,
            b: 0,
        },
        X {
            a: 0,
            b: 0,
        },
    ];

    function(&mut x);
}

You're not mutating SOME_CONST, as that ceases to exist at runtime. You're mutating the compiler-generated x variable.

const is basically a way of saying, run this code at compile time, then copy and paste the output whereever the name is used.

12

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