rust - 不能可变地借用模式守卫 [E0301]

Question

unsafe fn get_sync_handle(pid: u32) -> Option<HANDLE>
{
    let raw_h = OpenProcess(SYNCHRONIZE, 0, pid as DWORD);
    match raw_h
    {
        x if x >= 0 as HANDLE => None,
        x => Some(x)
    }
}

src\lib.rs:411:19: 411:23 error: cannot mutably borrow in a pattern guard [E0301] src\lib.rs:411 x if x >= zero => None

有人可以向我解释这个错误的原因吗？虽然我知道这是一个非常人为的例子，但我不明白为什么编译器会提示，我并没有试图修改任何东西，而且 raw_h 本身是不可变的。

Answer 1

这是一个独立的例子:

type Handle = *mut ();

fn main() {
    let foo = 0 as Handle;
    match foo {
        x if x >= 0 as Handle => None,
        x                     => Some(x)
    };
}

记下错误信息和它指向的地方:

<anon>:6:19: 6:30 error: cannot mutably borrow in a pattern guard [E0301]
<anon>:6         x if x >= 0 as Handle => None,
                           ^~~~~~~~~~~

其实是被吐槽的“null”。你可以翻转支票，然后比赛守卫中就不会有任何可变性。

type Handle = *mut ();

fn main() {
    let foo = 0 as Handle;
    match foo {
        x if x as isize >= 0 => None,
        x                    => Some(x)
    };
}