Function std::mem::take

pub fn take<T>(dest: &mut T) -> T where
    T: Default, 

🔬 This is a nightly-only experimental API. (mem_take #61129)

Replace dest with the default value of T, and return the previous dest value.

Examples

A simple example:

#![feature(mem_take)]

use std::mem;

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

let old_v = mem::take(&mut v);
assert_eq!(vec![1, 2], old_v);
assert!(v.is_empty());

take allows taking ownership of a struct field by replacing it with an "empty" value. Without take you can run into issues like these:

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 take can be used to disassociate the original value of self.buf from self, allowing it to be returned:

#![feature(mem_take)]

use std::mem;

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

let mut buffer = Buffer { buf: vec![0, 1] };
assert_eq!(buffer.buf.len(), 2);

assert_eq!(buffer.get_and_reset(), vec![0, 1]);
assert_eq!(buffer.buf.len(), 0);