1.0.0[−][src]Function std::mem::replace
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!(vec![1, 2], old_v); assert_eq!(vec![3, 4, 5], v);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 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 } }Run
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);Run