Function std::mem::replace1.0.0[][src]

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.

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!(2, old_v.len());
assert_eq!(3, v.len());Run

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

This example deliberately fails to compile
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
    }
}Run

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:

use std::mem;

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