It's always worth taking a look at the full compiler output:
error[E0308]: mismatched types
--> src/main.rs:39:20
|
39 | a.add_b(B::new(|| println!("test2")));
| ^^^^^^^^^^^^^^^^^^^^ expected closure, found a different closure
|
= note: expected type `[closure@src/main.rs:38:20: 38:39]`
found type `[closure@src/main.rs:39:20: 39:40]`
= note: no two closures, even if identical, have the same type
= help: consider boxing your closure and/or using it as a trait object
Especially helpful:
no two closures, even if identical, have the same type
consider boxing your closure and/or using it as a trait object
We can simplify your example further by removing the type B
altogether. Then the only task is to save a vector of closures. As the compiler tells us, no two closures have the same type, but Vec
is a homogeneous data structure, meaning that every item in it has the same type.
We can work around that restriction by introducing one level of indirection. As the compiler suggests, this can either be done by trait objects or boxing (the latter kind of includes the first one). The corresponding types would look like this:
Vec<&Fn()>
(reference to trait objects)
Vec<Box<Fn()>>
(trait object in a box)
In your example you want to own all closures, thus the correct choice is to box all closures, as Box<T>
is an owning wrapper while references only borrow stuff.
A fully working example:
struct A {
b_vec: Vec<B>,
}
impl A {
fn new() -> A {
A { b_vec: Vec::new() }
}
fn add_b(&mut self, b: B) {
self.b_vec.push(b);
}
}
struct B {
f: Box<Fn()>,
}
impl B {
fn new<F>(f: F) -> B
where
F: Fn() + 'static,
{
B { f: Box::new(f) }
}
}
fn main() {
let mut a = A::new();
a.add_b(B::new(|| println!("test")));
a.add_b(B::new(|| println!("test2")));
}
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