这是我的代码
struct test_loop {
is_break: bool,
}
impl test_loop {
fn run_forever<F>(&mut self, mut f: F)
where
F: FnMut() -> (),
{
self.is_break = false;
loop {
f();
if self.is_break {
break;
}
}
}
fn breakit(&mut self) {
self.is_break = true;
}
}
fn main() {
let mut l = test_loop { is_break: false };
let f = || {
l.breakit();
};
l.run_forever(f);
}
代码很简单,我不想在闭包中跳过调用l.breakit()
。现在编译器告诉我有第二个可变借用问题:
error[E0499]: cannot borrow `l` as mutable more than once at a time
--> src/main.rs:26:5
|
23 | let f = || {
| -- first mutable borrow occurs here
24 | l.breakit();
| - first borrow occurs due to use of `l` in closure
25 | };
26 | l.run_forever(f);
| ^ - first borrow later used here
| |
| second mutable borrow occurs here
我使用RefCell
来修复编译问题,但线程在运行时仍然会发生混乱。我应该删除关闭中的l.xxx
吗?或者有一些方法可以让代码像C ++或其他语言一样运行?
如果f()
可以改变test_loop
状态,那么将这个引用添加到它的签名是很自然的。这解决了第二个借用问题。
fn run_forever<F>(&mut self, mut f: F)
where
F: FnMut(&mut Self) -> (),
{
// call f(self) instead of f()
}
// main
let f = |l: &mut test_loop| {
l.breakit();
};