我正在学习/试验Rust,在我用这种语言找到的所有优雅中,有一个让我感到困惑并且看起来完全不合适的特点.
在进行方法调用时,Rust会自动取消引用指针.我做了一些测试来确定确切的行为:
struct X { val: i32 }
impl std::ops::Deref for X {
type Target = i32;
fn deref(&self) -> &i32 { &self.val }
}
trait M { fn m(self); }
impl M for i32 { fn m(self) { println!("i32::m()"); } }
impl M for X { fn m(self) { println!("X::m()"); } }
impl M for &X { fn m(self) { println!("&X::m()"); } }
impl M for &&X { fn m(self) { println!("&&X::m()"); } }
impl M for &&&X { …Run Code Online (Sandbox Code Playgroud) 鉴于此代码:
trait Base {
fn a(&self);
fn b(&self);
fn c(&self);
fn d(&self);
}
trait Derived : Base {
fn e(&self);
fn f(&self);
fn g(&self);
}
struct S;
impl Derived for S {
fn e(&self) {}
fn f(&self) {}
fn g(&self) {}
}
impl Base for S {
fn a(&self) {}
fn b(&self) {}
fn c(&self) {}
fn d(&self) {}
}
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不幸的是,我不能投&Derived给&Base:
fn example(v: &Derived) {
v as &Base;
}
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error[E0605]: non-primitive cast: `&Derived` as `&Base`
--> …Run Code Online (Sandbox Code Playgroud)