Why does “move” in Rust not actually move?

问题

In the below example:

struct Foo {
    a: [u64; 100000],
}

fn foo(mut f: Foo) -> Foo {
    f.a[0] = 99999;
    f.a[1] = 99999;
    println!("{:?}", &mut f as *mut Foo);

    for i in 0..f.a[0] {
        f.a[i as usize] = 21444;
    }

    return f;
}
fn main(){
    let mut f = Foo {
        a:[0;100000]
    };

    println!("{:?}", &mut f as *mut Foo);
    f = foo(f);
    println!("{:?}", &mut f as *mut Foo);
}

I find that before and after passing into the function foo, the address of f is different. Why does Rust copy such a big struct everywhere but not actually move it (or achieve this optimization)?

I understand how stack memory works. But with the information provided by ownership in Rust, I think the copy can be avoided. The compiler unnecessarily copies the array twice. Can this be an optimization for the Rust compiler?

回答1:

A move is a memcpy followed by treating the source as non-existent.

Your big array is on the stack. That's just the way Rust's memory model works: local variables are on the stack. Since the stack space of foo is going away when the function returns, there's nothing else the compiler can do except copy the memory to main's stack space.

In some cases, the compiler can rearrange things so that the move can be elided (source and destination are merged into one thing), but this is an optimization that cannot be relied on, especially for big things.

If you don't want to copy the huge array around, allocate it on the heap yourself, either via a Box<[u64]>, or simply by using Vec<u64>.

来源：https://stackoverflow.com/questions/53465843/why-does-move-in-rust-not-actually-move