线程生命周期错误

Question

我正在尝试使用Rust中的并发来实现游戏"River of Hanoi".老实说,我最后一次尝试了解Rust的整个生命周期,但我还没有完成.这就是为什么我得到一些我不理解的奇怪的生命周期错误.首先,这是重要的代码

fn move_plate<'a>(stack_a: &'a mut Vec<i32>, stack_b: &'a mut Vec<i32>, 
    stack_c: &'a mut Vec<i32>, moves: &'a mut Vec<(i32, i32)>) 
{
        let mut moves1: Vec<(i32, i32)> = Vec::new();
        let guard1 = Thread::scoped(
            move || { move_plate(stack_a, stack_c, stack_b, (1, 3, 2), &mut moves1); 
        });
        guard1.join().ok();
}

这是错误

error: cannot infer an appropriate lifetime due to conflicting requirements
    let guard1 = Thread::scoped(move || {
                     move_plate(height - 1, stack_a, stack_c, stack_b, (1, 3, 2), threads, depth + 1, &mut moves1);
                 });
note: first, the lifetime cannot outlive the expression at 93:25...
             let guard1 = Thread::scoped(move || {

note: ...so that the declared lifetime parameter bounds are satisfied
             let guard1 = Thread::scoped(move || {

note: but, the lifetime must be valid for the expression at 93:45...
             let guard1 = Thread::scoped(move || {
                 move_plate(height - 1, stack_a, stack_c, stack_b, (1, 3, 2), threads, depth + 1, &mut moves1);
             });
note: ...so type `closure[]` of expression is valid during the expression
             let guard1 = Thread::scoped(move || {
                 move_plate(height - 1, stack_a, stack_c, stack_b, (1, 3, 2), threads, depth + 1, &mut moves1);
             });
error: declared lifetime bound not satisfied
             let guard1 = Thread::scoped(move || {

我明白我必须避免线程超过函数,因为否则对移动的引用将会消失.但既然我加入了这个帖子,那应该没问题,不应该吗？那时我错过了什么？如果有人可以帮助我,那真的很好,我只是习惯那种很酷(但很复杂)的东西

Answer 1

这是Rust类型系统的已知限制.目前Rust只允许在线程之间发送数据,只有当这些数据满足Sendbound时,并且Send暗示'static- 也就是说,可以跨线程边界发送的唯一引用是'static1.

有一个RFC部分解除了这个限制,允许'static跨任务发送非引用.我认为它已被接受,但它不是(这很奇怪).已经创建了支持此类事物的API(这可能是您感到困惑的原因),但语言尚未调整.