如果捕獲變量，如何將閉包作為 function 參數傳遞

Question

我正在嘗試使用 Rust 中的高階函數。我很掙扎，因為我想傳遞的參數 function 是一個捕獲值的閉包。

這是我的第一次嘗試（ 游樂場鏈接）：

/// Takes a value, n, doubles it, and then applies the function f
fn double_then_f(n:u64, f: fn(u64) -> u64) -> u64 {
    f(n * 2)
}

fn main() {
    // Simple f closure works just fine
    let example_1 = double_then_f(5, |n| n + 1);
    
    // More realistic closure doesn't work
    let dynamic_value = vec![1, 2, 3].iter().sum::<u64>();
    let example_2 = double_then_f(5, |n| n + dynamic_value);
}

這無法編譯

note: expected fn pointer `fn(u64) -> u64`
                 found closure `[closure@src/main.rs:12:38: 12:41]`
note: closures can only be coerced to `fn` types if they do not capture any variables

在閱讀了Fn特征和fn類型之間的區別之后，這是我的第二次嘗試：

/// Takes a value, n, doubles it, and then applies the function f
fn double_then_f(n:u64, f: Box<dyn Fn(u64) -> u64>) -> u64 {
    f(n * 2)
}

fn main() {
    // Simple f closure works just fine
    let example_1 = double_then_f(5, Box::new(|n| n + 1));
    
    // More realistic closure doesn't work
    let dynamic_value = vec![1, 2, 3].iter().sum::<u64>();
    let example_2 = double_then_f(5, Box::new(|n| n + dynamic_value));
}

這也無法編譯，並出現與生命周期相關的錯誤，即“cast requires that dynamic_value is borrowed for 'static ”。 我想它正在談論的演員表是將我的閉包封裝到特征 object 中嗎？

我該怎么做才能使此代碼正常工作？

Answer 1

使用 Fn-traits！

fn double_then_f<F: Fn(u64) -> u64>(n:u64, f: F) -> u64 {
    f(n * 2)
}

fn main() {
    let example_1 = double_then_f(5, |n| n + 1);
    
    let dynamic_value = vec![1, 2, 3].iter().sum::<u64>();
    let example_2 = double_then_f(5, |n| n + dynamic_value);
}