`std :: mem :: swap`如何工作？

Question

在相同类型的两个可变位置交换值，而不进行初始化或复制任何一个。

use std::mem; 

let x = &mut 5; 
let y = &mut 42; 

mem::swap(x, y); 

assert_eq!(42, *x); 
assert_eq!(5, *y); 

（从offical Rust doc）

如何可以在两个值，而不拷贝被交换？价值42从y变为x是怎么回事？这应该是不可能的。

Answer 1

功能并不实际进行复印内部：这里是它的源从文档中提取：

pub fn swap<T>(x: &mut T, y: &mut T) { 
    unsafe { 
     // Give ourselves some scratch space to work with 
     let mut t: T = uninitialized(); 

     // Perform the swap, `&mut` pointers never alias 
     ptr::copy_nonoverlapping(&*x, &mut t, 1); 
     ptr::copy_nonoverlapping(&*y, x, 1); 
     ptr::copy_nonoverlapping(&t, y, 1); 

     // y and t now point to the same thing, 
     // but we need to completely forget `t` 
     // because it's no longer relevant. 
     forget(t); 
    } 
}