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。
最佳答案
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)
};
}
关于rust - 不能可变地借用模式守卫 [E0301],我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/30197745/