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我意识到 Rust 在不断变化，但无论如何我都在努力学习它。我试图了解如何将以下适用于 0.9 的示例改编为适用于 0.10 的类似示例：

fn main() {
    let argv = std::os::args();

    let (first, last) = match argv {
        [_, first_arg, .., last_arg] => (first_arg, last_arg),
        _ => fail!("Error: At least two arguments expected.")
    };

    println!("The first argument was {:s}, \
              and the last argument was {:s}.", first, last); 
}

当我用 0.10 构建它时，我收到以下错误：

error: couldn't read test.rc: no such file or directory (No such file or directory)
orflongpmacx8:rust pohl_longsine$ rustc test.rs
test.rs:9:9: 9:37 error: unique vector patterns are no longer supported
test.rs:9         [_, first_arg, .., last_arg] => (first_arg, last_arg),
                  ^~~~~~~~~~~~~~~~~~~~~~~~~~~~
error: aborting due to previous error

我的问题：是否仍然可以在 argv 上使用模式匹配，但使用不同的语法，或者根本不再可能在 argv 上使用 match 语句？如果是前者，我需要改变什么？

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1 回答 1

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You can still match on slices &[T] and fixed length arrays [T, .. n]. so, in this case,

fn main() {
    let argv = std::os::args();

    let (first, last) = match argv.as_slice() {
        [_, ref first_arg, .., ref last_arg] => (first_arg, last_arg),
        _ => fail!("Error: At least two arguments expected.")
    };

    println!("The first argument was {:s}, \
              and the last argument was {:s}.", *first, *last); 
}

Note the addition of the refs. argv is ~[~str], i.e. the contents are owned strings ~str, which move ownership when passed by value as a [_, first_arg, .., last_arg] pattern would do. It's illegal to move ownership out from behind a borrowed pointer (like a slice &[~str]) so that pattern would be illegal. One can borrow into the slice (and any other pattern) using the ref keyword, making first and last both references of type &~str.

One might wonder why the *first and *last dereferences aren't trying to move a ~str out from behind a &~str, but it's because the println! macro is expanding to essentially &*first and &*last, which is fine.

(We could also write => (first_arg.as_slice(), last_arg.as_slice()) to borrow the two &~strs straight to string slices &str, which then means we don't need to dereference in the println!.)

于 2014-04-05T00:04:28.250 回答