# Function std::mem::replace 1.0.0[−][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

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