mem::{take(_), replace(_)} to keep owned values in changed enums

Description

Say we have a &mut MyEnum which has (at least) two variants, A { name: String, x: u8 } and B { name: String }. Now we want to change MyEnum::A to a B if x is zero, while keeping MyEnum::B intact.

We can do this without cloning the name.

Example

#![allow(unused)]
fn main() {
use std::mem;

enum MyEnum {
    A { name: String, x: u8 },
    B { name: String },
}

fn a_to_b(e: &mut MyEnum) {
    if let MyEnum::A { name, x: 0 } = e {
        // This takes out our `name` and puts in an empty String instead
        // (note that empty strings don't allocate).
        // Then, construct the new enum variant (which will
        // be assigned to `*e`).
        *e = MyEnum::B {
            name: mem::take(name),
        }
    }
}
}

This also works with more variants:

#![allow(unused)]
fn main() {
use std::mem;

enum MultiVariateEnum {
    A { name: String },
    B { name: String },
    C,
    D,
}

fn swizzle(e: &mut MultiVariateEnum) {
    use MultiVariateEnum::*;
    *e = match e {
        // Ownership rules do not allow taking `name` by value, but we cannot
        // take the value out of a mutable reference, unless we replace it:
        A { name } => B {
            name: mem::take(name),
        },
        B { name } => A {
            name: mem::take(name),
        },
        C => D,
        D => C,
    }
}
}

Motivation

When working with enums, we may want to change an enum value in place, perhaps to another variant. This is usually done in two phases to keep the borrow checker happy. In the first phase, we observe the existing value and look at its parts to decide what to do next. In the second phase we may conditionally change the value (as in the example above).

The borrow checker won’t allow us to take out name of the enum (because something must be there.) We could of course .clone() name and put the clone into our MyEnum::B, but that would be an instance of the Clone to satisfy the borrow checker anti-pattern. Anyway, we can avoid the extra allocation by changing e with only a mutable borrow.

mem::take lets us swap out the value, replacing it with its default value, and returning the previous value. For String, the default value is an empty String, which does not need to allocate. As a result, we get the original name as an owned value. We can then wrap this in another enum.

NOTE: mem::replace is very similar, but allows us to specify what to replace the value with. An equivalent to our mem::take line would be mem::replace(name, String::new()).

Note, however, that if we are using an Option and want to replace its value with a None, Option’s take() method provides a shorter and more idiomatic alternative.

Advantages

Look ma, no allocation! Also you may feel like Indiana Jones while doing it.

Disadvantages

This gets a bit wordy. Getting it wrong repeatedly will make you hate the borrow checker. The compiler may fail to optimize away the double store, resulting in reduced performance as opposed to what you’d do in unsafe languages.

Furthermore, the type you are taking needs to implement the Default trait. However, if the type you’re working with doesn’t implement this, you can instead use mem::replace.

Discussion

This pattern is only of interest in Rust. In GC’d languages, you’d take the reference to the value by default (and the GC would keep track of refs), and in other low-level languages like C you’d simply alias the pointer and fix things later.

However, in Rust, we have to do a little more work to do this. An owned value may only have one owner, so to take it out, we need to put something back in – like Indiana Jones, replacing the artifact with a bag of sand.

See also

This gets rid of the Clone to satisfy the borrow checker anti-pattern in a specific case.