rust - 为什么在一个匹配项中使用未初始化的值时不会出现错误，而在使用 `panic`的另一个匹配项中却未出现错误？

Question

const FILE_NAME: &str = "/tmp/main.rs";

fn main() {
    println!("Hello, world!");
    let metadata;
    match std::fs::metadata(FILE_NAME) {
        Ok(meta_data) => {
            metadata = meta_data;
            println!("metadata: {:?}", metadata);
        }
        _ => {
            panic!("not meta enough!");
        }
    }

    let modified: std::time::SystemTime;
    match metadata.modified() {
        Ok(x) => {
            modified = x;
            println!("modified: {:?}", modified); // this works
        }
        _ => (),
    }

    // println!("modified: {:?}", modified); //-> this does not work
    let len = metadata.len();
    println!("metadata: {:?}", metadata);
    println!("len: {}", len);
}

我不明白如果取消注释第22行，为什么它失败了:

error[E0381]: borrow of possibly-uninitialized variable: `modified`
  --> src/main.rs:25:32
   |
25 |     println!("modified: {:?}", modified); //-> this does not work
   |                                ^^^^^^^^ use of possibly-uninitialized `modified`

第18行的相同代码可以正常工作。我在俯视什么？

Answer 1

在有问题的行中，Rust无法确定modified是否已实际初始化。

如果无法设置modified，则可以选择提前返回:

fn main() {
    println!("Hello, world!");
    let metadata;
    match std::fs::metadata("dsf") {
        Ok(meta_data) => {
            metadata = meta_data;
            println!("metadata: {:?}", metadata);
        }
        _ => {
            panic!("not meta enough!");
        }
    }

    let modified: std::time::SystemTime = match metadata.modified() {
        Ok(x) => {
            println!("modified is set to {:?}", x); // this works
            x
        }
        _ => return, // early return on error
    };

    println!("modified: {:?}", modified);
    let len = metadata.len();
    println!("metadata: {:?}", metadata);
    println!("len: {}", len);
}