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964dd99 Jan 1, 2017
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@llogiq @partim @diwic @lfairy
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mem::replace to keep owned values in changed enums


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.


use std::mem;

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

fn a_to_b(e: &mut MyEnum) {

    // we mutably borrow `e` here. This precludes us from changing it directly
    // as in `*e = ...`, because the borrow checker won't allow it. Therefore
    // the assignment to `e` must be outside the `if let` clause. 
    *e = if let MyEnum::A { ref mut name, x: 0 } = *e {
        // this takes out our `name` and put in an empty String instead
        // (note that empty strings don't allocate).
        // Then, construct the new enum variant (which will 
        // be assigned to `*e`, because it is the result of the `if let` expression).
        MyEnum::B { name: mem::replace(name, String::new()) }
    // In all other cases, we return immediately, thus skipping the assignment
    } else { return }

This also works with more variants:

use std::mem;

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

fn swizzle(e: &mut MultiVariateEnum) {
    use self::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 { ref mut name } => B { name: mem::replace(name, String::new()) },
        B { ref mut name } => A { name: mem::replace(name, String::new()) },
        C => D,
        D => C


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] antipattern. Anyway, we can avoid the extra allocation by changing e with only a mutable borrow.

mem::replace lets us swap out the value, replacing it with something else. In this case, we put in 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, 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.


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


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.


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] antipattern in a specific case.

[Clone to satisfy the borrow checker](TODO: Hinges on PR #23)