Question

这是一个compose函数，可以组成Fn个闭包：

fn compose<'a, T1, T2, T3, F1, F2>(f: F1, g: F2) -> Box<Fn(T1) -> T3 + 'a>
    where F1: Fn(T1) -> T2 + 'a,
          F2: Fn(T2) -> T3 + 'a
{
    box move |x| g(f(x))
}

如何使这个撰写函数可以关闭FnMut？我试过了：

fn compose<'a, T1, T2, T3, F1, F2>(f: F1, g: F2) -> Box<FnMut(T1) -> T3 + 'a>
    where F1: FnMut(T1) -> T2 + 'a,
          F2: FnMut(T2) -> T3 + 'a
{
    box move |x| g(f(x))
}

但它抱怨：

error: cannot borrow captured outer variable in an `FnMut` closure as mutable
         box move |x| g(f(x))
                      ^
error: cannot borrow captured outer variable in an `FnMut` closure as mutable
         box move |x| g(f(x))
                        ^

扩展这一点，是否可以使用FnOnce闭包？

Answer 1

局部变量f和g必须是可变的：

fn compose<'a, T1, T2, T3, F1, F2>(mut f: F1, mut g: F2) -> Box<FnMut(T1) -> T3 + 'a>
    where F1: FnMut(T1) -> T2 + 'a,
          F2: FnMut(T2) -> T3 + 'a
{
    Box::new(move |x| g(f(x)))
}

如何编写可以组成`FnMut`闭包的函数？

1 个答案: