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[英]Rust: error[E0495]: cannot infer an appropriate lifetime for autorefdue to conflicting requirements in closure
[英]Closure as function parameter “cannot infer an appropriate lifetime due to conflicting requirements”
我正在嘗試使用閉包作為函數參數:
fn foo(f: Box<Fn() -> bool>) -> bool {
f()
}
fn main() {
let bar = 42;
foo(Box::new(|| bar != 42));
}
但我得到這個終身錯誤:
src/main.rs:7:24: 7:36 error: cannot infer an appropriate lifetime due to conflicting requirements
src/main.rs:7 let n = foo(Box::new(|| bar != 42));
^~~~~~~~~~~~
src/main.rs:7:15: 7:23 note: first, the lifetime cannot outlive the expression at 7:14...
src/main.rs:7 let n = foo(Box::new(|| bar != 42));
^~~~~~~~
src/main.rs:7:15: 7:23 note: ...so that the type `[closure src/main.rs:7:24: 7:36]` will meet its required lifetime bounds
src/main.rs:7 let n = foo(Box::new(|| bar != 42));
^~~~~~~~
src/main.rs:7:15: 7:37 note: but, the lifetime must be valid for the call at 7:14...
src/main.rs:7 let n = foo(Box::new(|| bar != 42));
^~~~~~~~~~~~~~~~~~~~~~
src/main.rs:7:24: 7:36 note: ...so that argument is valid for the call
src/main.rs:7 let n = foo(Box::new(|| bar != 42));
^~~~~~~~~~~~
error: aborting due to previous error
我不明白為什么不能正確推斷壽命。 我該怎么解決?
$ rustc --version
rustc 1.0.0-nightly (6c065fc8c 2015-02-17) (built 2015-02-18)
如果要使用盒裝閉合,則需要使用move || {}
move || {}
。
fn foo(f: Box<Fn() -> bool>)
-> bool {
f()
}
fn main() {
let bar = 42;
let blub = foo(Box::new(move || bar != 42));
}
另一方面,您不能直接使用未裝箱的封閉盒,因為它可能包含任何數量的捕獲元素,因此大小不一。 通過使用泛型,您可以輕松繞開此限制:
fn foo<T>(f: T)
-> bool
where T : Fn() -> bool {
f()
}
fn main() {
let bar = 42;
let blub = foo(|| bar != 42);
}
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