Question

我正在尝试编写一个函数，它接受一个闭包并返回任何闭包返回类型。我不想要硬编码的返回类型，比如 fn(action: &dyn Fn()) -> i32。我试图解决它，如下所示：

pub fn execute_code<T>(actions: T) -> T::Output
    where
        T: Fn(),
{
    println!("before:");
    let res = actions();
    println!("after:");
    res
}

fn main() {
    let i32_res: i32 = execute_code(&|| 5 + 5);
    let str_res: &str = execute_code(&|| "a string");
    println!("i32_res: {}, string_res: {}", i32_res, str_res);
}

但是编译器（1.53 稳定版）抱怨：

pub fn execute_code<T>(actions: T) -> T::Output
   |        ------------ required by a bound in this
2  |     where
3  |         T: Fn(),
   |            ---- required by this bound in `execute_code`
...
12 |     let i32_res: i32 = execute_code(&|| 5 + 5);
   |                        ^^^^^^^^^^^^ expected `()`, found integer

实现这一目标的正确方法是什么？

Answer 1

您只需要将输出参数化为另一种类型：

pub fn execute_code<T, R>(actions: T) -> R
    where
        T: Fn() -> R,
{
    println!("before:");
    let res = actions();
    println!("after:");
    res
}

fn main() {
    let i32_res: i32 = execute_code(&|| 5 + 5);
    let str_res: &str = execute_code(&|| "a string");
    println!("i32_res: {}, string_res: {}", i32_res, str_res);
}

Playground

在 Rust 中编写一个函数，它接受闭包/函数 `T` 并返回 `T::Output`

1 个答案: