[英]How can I use a function as a closure when the function needs to take a reference to the closures argument?
[英]How can I pass a FnMut closure to a function using a reference in Rust?
我已经学会了如何将一个闭包参数传递给一个函数,所以我可以调用两次closure
:
let closure = || println!("hello");
fn call<F>(f: &F)
where
F: Fn(),
{
f();
}
call(&closure);
call(&closure);
当我使用FnMut
:
let mut string: String = "hello".to_owned();
let change_string = || string.push_str(" world");
fn call<F>(mut f: &mut F)
where
F: FnMut(),
{
f();
}
call(&change_string);
call(&change_string);
结果会出错:
error[E0308]: mismatched types
--> src/main.rs:10:10
|
10 | call(&change_string);
| ^^^^^^^^^^^^^^ types differ in mutability
|
= note: expected type `&mut _`
found type `&[closure@src/main.rs:3:25: 3:53 string:_]`
我该如何解决?
正如错误消息所示:
expected type `&mut _`
found type `&[closure@src/main.rs:3:25: 3:53 string:_]`
它期待对某些东西 ( &mut _
)的可变引用,但是你提供了一个对闭包( &...
)的不可变引用。 采取可变参考:
call(&mut change_string);
这会导致下一个错误:
error: cannot borrow immutable local variable `change_string` as mutable
--> src/main.rs:9:15
|
3 | let change_string = || string.push_str(" world");
| ------------- use `mut change_string` here to make mutable
...
9 | call(&mut change_string);
| ^^^^^^^^^^^^^ cannot borrow mutably
采用可变引用要求值本身是可变的:
let mut change_string = || string.push_str(" world");
在这种情况下,您根本不需要使用&mut F
,因为FnMut
是针对FnMut
的可变引用FnMut
。 也就是说,这有效:
fn call(mut f: impl FnMut()) {
f();
}
call(&mut change_string);
call(&mut change_string);
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