Function core::mem::replace

pub fn replace<T>(dest: &mut T, src: T) -> T

Replaces the value at a mutable location with a new one, returning the old value, without deinitializing or copying either one.

This is primarily used for transferring and swapping ownership of a value in a mutable location.

Examples

A simple example:

fn main() { use std::mem; let mut v: Vec<i32> = Vec::new(); mem::replace(&mut v, Vec::new()); }

use std::mem;

let mut v: Vec<i32> = Vec::new();

mem::replace(&mut v, Vec::new());

This function allows consumption of one field of a struct by replacing it with another value. The normal approach doesn't always work:

fn main() { struct Buffer<T> { buf: Vec<T> } impl<T> Buffer<T> { fn get_and_reset(&mut self) -> Vec<T> { // error: cannot move out of dereference of `&mut`-pointer let buf = self.buf; self.buf = Vec::new(); buf } } }

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

impl<T> Buffer<T> {
    fn get_and_reset(&mut self) -> Vec<T> {
        // error: cannot move out of dereference of `&mut`-pointer
        let buf = self.buf;
        self.buf = Vec::new();
        buf
    }
}

Note that T does not necessarily implement Clone, so it can't even clone and reset self.buf. But replace can be used to disassociate the original value of self.buf from self, allowing it to be returned:

fn main() { #![allow(dead_code)] use std::mem; struct Buffer<T> { buf: Vec<T> } impl<T> Buffer<T> { fn get_and_reset(&mut self) -> Vec<T> { mem::replace(&mut self.buf, Vec::new()) } } }

use std::mem;
impl<T> Buffer<T> {
    fn get_and_reset(&mut self) -> Vec<T> {
        mem::replace(&mut self.buf, Vec::new())
    }
}