I Want Off Mr. Golang's Wild Ride

Re:I Want Off Mr. Golang's Wild Ride by joshuamorton on Friday February 28, 2020 @03:29PM (#22446035)

Well no, for the reason I describe above: if you have the pattern

    mut a = 4
    f(a)
    print(a)

In rust and python, you'll always get 4 in python, but the value in rust depends on `f`.

This means that the passed variable is immutable but shadowable, as in rust. (An object in python is much more like an Box/Cell, so the contained object can be mutated, but the reference to the box itself is immutable).

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Re:I Want Off Mr. Golang's Wild Ride by hope-striker on Friday February 28, 2020 @03:32PM (#22446064)

The value in Rust does not depend on f.

I will be precise. There is no definition of f such that this function will print anything other than "4".

    fn main() {
        let mut x = 4;
        f(x);
        println!("{}", x);
    }

Again, you seem to be confusing mutable references and mutable variables. If I had written f(&mut x) rather than f(x), you would be right.

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