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Function std::mem::take

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

Replaces dest with the default value of T, returning the previous dest value.

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

Examples

A simple example:

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:

ⓘ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
    }
}

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:

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);

© 2010 The Rust Project Developers
Licensed under the Apache License, Version 2.0 or the MIT license, at your option.
https://doc.rust-lang.org/std/mem/fn.take.html