我真的不知道该怎么问这个问题，但有人能解释一下为什么会这样吗？
https://play.rust-lang.org/?version=stable&mode=debug&edition=2018&gist=969cf50f66746c4aa3941200e01f1570

enum SlotFn<'a, Comp, Args = (), R = ()> 
where Args: 'static,
R: Default + 'static,
{
    SimpleFn(fn(Args) -> R),
    MemberFn(fn(&'a Comp) -> R),
    MemberFnMut(fn(&'a mut Comp, Args) -> R),
}

impl<'a, Comp, Args, R> SlotFn<'a, Comp, Args, R> 
where Args: 'static,
R: Default + 'static,{
    fn from_member(f: fn(&'a Comp) -> R) -> Self{
        SlotFn::MemberFn(f)
    }

    fn from_member_mut(f: fn(&'a mut Comp, Args) -> R) -> Self {
        SlotFn::MemberFnMut(f)
    }

    fn emit(&self, comp: &'a Comp, args: Args) -> R {
        match self {
            SlotFn::SimpleFn(f) => f(args),
            SlotFn::MemberFn(f) => f(comp),
            _ => Default::default()
        }
    }

    fn emit_mut(&mut self, comp: &'a mut Comp, args: Args) -> R {
        match self {
            SlotFn::MemberFnMut(f) => f(comp, args),
            _ => Default::default()
        }
    }
}

struct Test(u32);

impl Test {

    fn reffunc(&self) {
        println!("value: {}", self.0);
    }

    fn reffunc_mut(&mut self, val: u32) {
        self.0 = val;
    }
}

fn main() {
    let mut test = Test(0);
    let slot = SlotFn::from_member(Test::reffunc);
    let mut mslot = SlotFn::from_member_mut(Test::reffunc_mut);

    mslot.emit_mut(&mut test, 10);
    slot.emit(&test, ());
}

但不是这个
https://play.rust-lang.org/?version=stable&mode=debug&edition=2018&gist=1534d9b49327272ddbbd04826efd644a

enum SlotFn<'a, Comp, Args = (), R = ()> 
where Args: 'static,
R: Default + 'static,
{
    SimpleFn(Box<dyn Fn(Args) -> R + Send + Sync + 'static>),
    MemberFn(Box<dyn Fn(&'a Comp) -> R + Send + Sync + 'static>),
    MemberFnMut(Box<dyn FnMut(&'a mut Comp, Args) -> R + Send + Sync + 'static>),
}

impl<'a, Comp, Args, R> SlotFn<'a, Comp, Args, R> 
where Args: 'static,
R: Default + 'static,{
  
    fn from_member<F>(f: F) -> Self where F: Fn(&'a Comp) -> R + Send + Sync + 'static{
        SlotFn::MemberFn(Box::new(f))
    }

    fn from_member_mut<F>(f: F) -> Self where F: FnMut(&'a mut Comp, Args) -> R + Send + Sync + 'static{
        SlotFn::MemberFnMut(Box::new(f))
    }

    fn emit(&self, comp: &'a Comp, args: Args) -> R {
        match self {
            SlotFn::SimpleFn(f) => f(args),
            SlotFn::MemberFn(f) => f(comp),
            _ => Default::default()
        }
    }

    fn emit_mut(&mut self, comp: &'a mut Comp, args: Args) -> R {
        match self {
            SlotFn::MemberFnMut(f) => f(comp, args),
            _ => Default::default()
        }
    }
}

struct Test(u32);

impl Test {

    fn reffunc(&self) {
        println!("value: {}", self.0);
    }

    fn reffunc_mut(&mut self, val: u32) {
        self.0 = val;
    }
}

fn main() {
    let mut test = Test(0);
    let slot = SlotFn::from_member(Test::reffunc);
    let mut mslot = SlotFn::from_member_mut(Test::reffunc_mut);

    mslot.emit_mut(&mut test, 10);
    slot.emit(&test, ());
}

在第二种情况下，我得到了以下错误：

error[E0502]: cannot borrow `test` as immutable because it is also borrowed as mutable

mutable borrow might be used here, when `mslot` is dropped and runs the destructor for type `SlotFn<'_, Test, u32>`