封闭物，函数参数“不能推断适当的寿命因相互矛盾的要求”(Closure as function pa

我想使用闭包的功能参数：

fn foo(f: Box<Fn() -> bool>) -> bool {
    f()
}

fn main() {
    let bar = 42;
    foo(Box::new(|| bar != 42));
}

但我得到这个一生的错误：

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

我不明白为什么寿命不正确infered。 我能做些什么来解决这个问题？

$ rustc --version
rustc 1.0.0-nightly (6c065fc8c 2015-02-17) (built 2015-02-18)

Answer 1:

if you want to use a boxed closure you need to use move || {}.

fn foo(f: Box<Fn() -> bool>)
       -> bool {
    f()
}

fn main() {
    let bar = 42;
    let blub = foo(Box::new(move || bar != 42));
}

On the other hand, you cannot use an unboxed closure directly, as it may contain any number of captured elements, and is therefor unsized. By using generics you can easily circumvent this limitation:

fn foo<T>(f: T)
          -> bool
          where T : Fn() -> bool {
    f()
}

fn main() {
    let bar = 42;
    let blub = foo(|| bar != 42);
}

文章来源: Closure as function parameter “cannot infer an appropriate lifetime due to conflicting requirements”