Function scale_info::prelude::mem::replace

#[must_use =
  "if you don't need the old value, you can just assign the new value directly"]pub fn replace<T>(dest: &mut T, src: T) -> T

Moves src into the referenced dest, returning the previous dest value.

Neither value is dropped.

• If you want to replace the values of two variables, see swap.
• If you want to replace with a default value, see take.

Examples

A simple example:

use std::mem;

let mut v: Vec<i32> = vec![1, 2];

let old_v = mem::replace(&mut v, vec![3, 4, 5]);
assert_eq!(vec![1, 2], old_v);
assert_eq!(vec![3, 4, 5], v);

replace allows consumption of a struct field by replacing it with another value. Without replace you can run into issues like these:

struct Buffer<T> { buf: Vec<T> }

impl<T> Buffer<T> {
    fn replace_index(&mut self, i: usize, v: T) -> T {
        // error: cannot move out of dereference of `&mut`-pointer
        let t = self.buf[i];
        self.buf[i] = v;
        t
    }
}

Note that T does not necessarily implement Clone, so we can't even clone self.buf[i] to avoid the move. But replace can be used to disassociate the original value at that index from self, allowing it to be returned:

use std::mem;

impl<T> Buffer<T> {
    fn replace_index(&mut self, i: usize, v: T) -> T {
        mem::replace(&mut self.buf[i], v)
    }
}

let mut buffer = Buffer { buf: vec![0, 1] };
assert_eq!(buffer.buf[0], 0);

assert_eq!(buffer.replace_index(0, 2), 0);
assert_eq!(buffer.buf[0], 2);