Rust总结：可恢复的错误Result

by matt Leave Comment

可恢复的错误 Result

unwrap/expect 如果失败就panic

use std::fs::File;

fn main() {
    // 如果调用这段代码时 hello.txt 文件不存在，那么 unwrap 就将直接 panic
    let f = File::open("hello.txt").unwrap();
}

? 和报错传播

fn open_file() -> Result<File, Box<dyn std::error::Error>> {
    // 如果失败就自动return Err
    let mut f = File::open("hello.txt")?;
    Ok(f)
}

? 不仅仅可以用于 Result 的传播，还能用于 Option 的传播

fn last_char_of_first_line(text: &str) -> Option<char> {
    text.lines().next()?.chars().last()
}

新手用 ? 常会犯的错误

初学者在用 ? 时，老是会犯错，例如写出这样的代码：

fn first(arr: &[i32]) -> Option<&i32> {
   arr.get(0)?
}

切记：? 操作符需要一个变量来承载正确的值，这个函数只会返回 Some(&i32) 或者 None，然而只有错误值能直接返回，正确的值不行

实际上 Rust 还支持另外一种形式的 main 函数

use std::error::Error;
use std::fs::File;

fn main() -> Result<(), Box<dyn Error>> {
    let f = File::open("hello.txt")?;

    Ok(())
}

例子

1

use std::num::ParseIntError;

fn multiply(n1_str: &str, n2_str: &str) -> Result<i32, ParseIntError> {
    let n1 = n1_str.parse::<i32>();
    let n2 = n2_str.parse::<i32>();
    Ok(n1.unwrap() * n2.unwrap())
}

fn main() {
    let result = multiply("10", "2");
    assert_eq!(result, Ok(20));

    let result = multiply("4", "2");
    assert_eq!(result.unwrap(), 8);

    println!("Success!")
}

用? 改写

use std::num::ParseIntError;

// IMPLEMENT multiply with ?
// DON'T use unwrap here
fn multiply(n1_str: &str, n2_str: &str) -> Result<i32, ParseIntError> {
    let n1 = n1_str.parse::<i32>()?;
    let n2 = n2_str.parse::<i32>()?;
    Ok(n1 * n2)
}

fn main() {
    assert_eq!(multiply("3", "4").unwrap(), 12);
    println!("Success!")
}

连续?

use std::fs::File;
use std::io::{self, Read};

fn read_file1() -> Result<String, io::Error> {
    let f = File::open("hello.txt");
    let mut f = match f {
        Ok(file) => file,
        Err(e) => return Err(e),
    };

    let mut s = String::new();
    match f.read_to_string(&mut s) {
        Ok(_) => Ok(s),
        Err(e) => Err(e),
    }
}

fn read_file2() -> Result<String, io::Error> {
    let mut s = String::new();
    // 注意 read_to_string(&mut s)?的?虽然不写也可以通过，但是就错过了可能报的错
    File::open("hello.txt")?.read_to_string(&mut s)?;

    Ok(s)
}

fn main() {
    assert_eq!(read_file1().unwrap_err().to_string(), read_file2().unwrap_err().to_string());
    println!("Success!")
}

map和and_then的使用

use std::num::ParseIntError;

fn add_two(n_str: &str) -> Result<i32, ParseIntError> {
   n_str.parse::<i32>().map(|num| num +2)
}

fn main() {
    assert_eq!(add_two("4").unwrap(), 6);

    println!("Success!")
}
use std::num::ParseIntError;

fn add_two(n_str: &str) -> Result<i32, ParseIntError> {
   n_str.parse::<i32>().and_then(|num| Ok(num +2))
}

fn main() {
    assert_eq!(add_two("4").unwrap(), 6);

    println!("Success!")
}

用map/and_then改写

use std::num::ParseIntError;

// With the return type rewritten, we use pattern matching without `unwrap()`.
// But it's so Verbose..
fn multiply(n1_str: &str, n2_str: &str) -> Result<i32, ParseIntError> {
    match n1_str.parse::<i32>() {
        Ok(n1)  => {
            match n2_str.parse::<i32>() {
                Ok(n2)  => {
                    Ok(n1 * n2)
                },
                Err(e) => Err(e),
            }
        },
        Err(e) => Err(e),
    }
}

// Rewriting `multiply` to make it succinct
// You  MUST USING `and_then` and `map` here
fn multiply1(n1_str: &str, n2_str: &str) -> Result<i32, ParseIntError> {
    // IMPLEMENT...
    n1_str.parse::<i32>().and_then(|n1| {
        n2_str.parse::<i32>().map(|n2| n1 * n2)
    })
}

fn print(result: Result<i32, ParseIntError>) {
    match result {
        Ok(n)  => println!("n is {}", n),
        Err(e) => println!("Error: {}", e),
    }
}

fn main() {
    // This still presents a reasonable answer.
    let twenty = multiply1("10", "2");
    print(twenty);

    // The following now provides a much more helpful error message.
    let tt = multiply("t", "2");
    print(tt);

    println!("Success!")
}

简化类型

use std::num::ParseIntError;

// Define a generic alias for a `Result` with the error type `ParseIntError`.
type Res<T> = Result<T, ParseIntError>;

// Use the above alias to refer to our specific `Result` type.
fn multiply(first_number_str: &str, second_number_str: &str) -> Res<i32> {
    first_number_str.parse::<i32>().and_then(|first_number| {
        second_number_str.parse::<i32>().map(|second_number| first_number * second_number)
    })
}

// Here, the alias again allows us to save some space.
fn print(result: Res<i32>) {
    match result {
        Ok(n)  => println!("n is {}", n),
        Err(e) => println!("Error: {}", e),
    }
}

fn main() {
    print(multiply("10", "2"));
    print(multiply("t", "2"));
}

main里的return问题

原题如果改成用特征对象就会报错

use std::num::ParseIntError;

fn main() -> Result<(), Box<dyn std::error::Error>> {
    let number_str = "10";
    let number = match number_str.parse::<i32>() {
        Ok(number)  => number,
        Err(e) => return Err(e),
    };
    println!("{}", number);
    Ok(())
}

用?就成功，这是为什么呢？不是说?是个宏，其实也是return了Err(e)的吗？

use std::num::ParseIntError;

fn main() -> Result<(), Box<dyn std::error::Error>> {
    let number_str = "10";
    let number = number_str.parse::<i32>()?;
    println!("{}", number);
    Ok(())
}

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