generics - 包含闭包参数的方法需要错误的类型？

Question

我有一个结构体，Parser其中包含一个 transformer - 用于修改其 ParserState 的函数- 和说ParserState .

#[derive(Clone)]
pub struct Parser<F> 
where
    F: Fn(ParserState) -> ParserState
{
    pub transformer: F,
    pub state: ParserState
}

#[derive(Debug, PartialEq, Clone)]
pub struct ParserState {
    pub target: String,
    pub index: usize,
    pub result: Vec<String>, // the container of eventual results from the parsing, with Some(result) or None. Should be a vetor
    pub error: bool,            // whether we've encountered an error; index -> 0, Some(err_msg)
    pub err_msg: Option<String> // Eventual error message
}

定义 .map()方法，我假设我可以简单地使用一个闭包，并在其中引用解析器自己的状态，将其作为参数传递给该闭包。此闭包将用作 .map() 的参数方法。

impl<F> Parser<F>
where
    F: Fn(ParserState) -> ParserState 
{
    pub fn new(f: F) -> Self {
        // creating a new Parser just means deciding on which closure it applies
        Parser {
            transformer: f,
            state: ParserState {
                target: String::from(""),
                index: 0,
                result: vec![],
                error: false,
                err_msg: None
            }
        }
    }

    pub fn map<G>(&mut self, g: G) -> ()
    where
        G: Fn(ParserState) -> ParserState
    {
        self.state = g((self.state).clone())
    }

    pub fn run(mut self, corpus: String) -> Self {
        self.state.target = corpus;
        self.state = (self.transformer)(self.state);
        self
    }
}

我是这样使用它的:

fn main() {
    let haystack: String = String::from("Hello!Goodbye!");
    let needle = String::from("Hello!");
    let str_parser = Parser::new(str_parser(needle));
    let closure = |mut state: ParserState| state.index = 0;
    let result = str_parser.run(haystack);
    let result = result.map(closure);
    let adv = ParserState {
        target: "Hello!Goodbye!".to_string(),
        index: 0,
        result: vec!["Hello!".to_string()],
        error: false,
        err_msg: None
    };
    assert_eq!(adv, result.state);
}

这没有奏效。我遇到了这个错误:

error[E0271]: type mismatch resolving `<[closure@src/lib.rs:49:23: 49:63] as std::ops::FnOnce<(parse
rs::ParserState,)>>::Output == parsers::ParserState`
  --> src/lib.rs:51:29
   |
51 |         let result = result.map(closure);
   |                             ^^^ expected `()`, found struct `parsers::ParserState`

并且不确定如何 FnOnce ，我没有提到，是相关的。我也不明白怎么 .map()预计 ()作为参数，当它是 Parser 的方法时.我觉得这里有点超出我的深度。

这是指向 the relevant Rust playground 的链接

Answer 1

此关闭返回 () (即它不返回值):

let closure = |mut state: ParserState| state.index = 0;

但是 F声明返回 ParserState .您只需要按预期使函数返回值即可。

let closure = |mut state: ParserState| {
    state.index = 0;
    state
};

and am not sure how FnOnce, which I have not mentioned, is relevant.

FnOnce是最普遍的功能特征；所有函数至少可以被调用一次，所以所有函数都实现了 FnOnce ，但可能不会实现 Fn或 FnMut .我认为它提到的事实 FnOnce在错误消息中只是它尝试匹配类型的顺序的一个工件。