匹配武器：“预期类型不匹配（），找到整型变量”

Question

我编写了以下代码来解析字符串以获取其中编码的整数，并使用match. 如果我得到一个Err(e)我想打印出错误e，并返回一个默认值。

return match t.parse::<usize>() {
    Ok(n) => n,
    Err(e) => {
        println!("Couldn't parse the value for gateway_threads {}", e);
        // Return two as a default value
        return 2;
    },
};

然而，该代码无法编译，因为它需要类型()但得到一个整数：

return match t.parse::<usize>() {
    Ok(n) => n,
    Err(e) => {
        println!("Couldn't parse the value for gateway_threads {}", e);
        // Return two as a default value
        return 2;
    },
};

如果我删除默认值的返回，我会收到错误expected usize but got `()`：

error[E0308]: mismatched types
  --> src/main.rs:37:32
   |
37 |                         return 2;
   |                                ^ expected (), found integral variable
   |
   = note: expected type `()`
              found type `{integer}

完整代码（我正在解析 INI 配置文件以获取一些值）：

extern crate threadpool;
extern crate ini;

use std::net::{TcpListener, TcpStream};
use std::io::Read;
use std::process;
use threadpool::ThreadPool;
use ini::Ini;

fn main() {

    let mut amount_workers: usize;
    let mut network_listen = String::with_capacity(21);
    //Load INI
    {
        let conf: Ini = match Ini::load_from_file("/etc/iotcloud/conf.ini") {
            Ok(t) => t,
            Err(e) => {
                println!("Error load ini file {}", e);
                process::exit(0);
            },
        };
        let section = match conf.section(Some("network".to_owned())) {
            Some(t) => t,
            None => {
                println!("Couldn't find the network ");
                process::exit(0);
            },
        };
        //amount_workers = section.get("gateway_threads").unwrap().parse().unwrap();
        amount_workers = match section.get("gateway_threads") {
            Some(t) => {
                return match t.parse::<usize>() {
                    Ok(n) => n,
                    Err(e) => {
                        println!("Couldn't parse the value for gateway_threads {}", e);
                        // Return two as a default value
                        return 2; //ERROR HERE;
                    },
                };
            },
            None => 2, // Return two as a default value
        };
        let ip = section.get("bind_ip").unwrap();
        let port = section.get("bind_port").unwrap();
        network_listen.push_str(ip);
        network_listen.push_str(":");
        network_listen.push_str(port);
    }
}

导致此错误的原因是什么？

Answer 1

改变

amount_workers = match section.get("gateway_threads") {
    Some(t) => {
        return match t.parse::<usize>() {
            Ok(n) => n,
            Err(e) => {
                println!("Couldn't parse the value for gateway_threads {}", e); //TODO: Log this
                return 2; //ERROR HERE; //Default value is set to 2
            }
        };
    }
    None => 2, //Default value is set to 2
};

到

amount_workers = match section.get("gateway_threads") {
    Some(t) => {
        match t.parse::<usize>() {  // No return
            Ok(n) => n,
            Err(e) => {
                println!("Couldn't parse the value for gateway_threads {}", e); //TODO: Log this
                2  // No semicolon, no return
            }
        } // No semicolon
    }
    None => 2, //Default value is set to 2
};

在 Rust 中，不以 结尾语句;是返回值的方式。return当您希望整个函数在最后一行之前返回一个值时，请使用该关键字，这就是您将其称为“早期返回”的原因 。

您可以在此处找到有关 Rust 如何处理表达式的更多信息。