# 1.40.0[−][src]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());

Run

`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
        }
    }

Run

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);

Run