rust - 注释生命周期/确定方法中字符串所有权的正确方法？

Question

只是从我在Rust中的冒险开始，并试图找出创建和返回枚举的正确方法，该枚举应该拥有通过传递给该方法的String创建的＆str字符串切片的所有权。我认为生命周期注释本身并不能解决问题，因为无论如何，原始字符串参数将超出功能块末尾的范围。任何帮助将非常感激。

pub enum PubSubMessage {
    // For now I won't worry about subbing/unsubbing to an array of channels
    SUBSCRIBE { channel: &str },
    UNSUBSCRIBE { channel: &str },
    PUBLISH { channel: &str, msg: &str},
    PSUBSCRIBE { pattern: &str },
    PUNSUBSCRIBE { pattern: &str},
}

pub fn parse_message(msg: String) -> Result<PubSubMessage, String> {
    let msg_contents: Vec<&str> = msg.split(" ").collect();

    return match msg_contents.as_slice() {
        ["SUBSCRBE", channel] => Ok(PubSubMessage::SUBSCRIBE { channel }),
        ["UNSUBSCRIBE", channel] => Ok(PubSubMessage::UNSUBSCRIBE { channel }),
        ["PUBLISH", channel, msg] => Ok(PubSubMessage::PUBLISH { channel, msg }),
        _ => Err("Could not parse ws message.".to_string())
    }
}

我当前遇到的编译器错误只是枚举定义需要生存期参数。

Answer 1

正如其他人在评论中所说的那样，最简单的方法是在各处使用String。话虽如此，如果您想避免多余的副本和分配，则可以将拥有该字符串的责任推向更高的调用链。如果要执行此操作，则需要更改函数签名，以使其借用msg而不是获得所有权，并添加所需的生存期参数。像这样的东西:

pub enum PubSubMessage<'a> {
    // For now I won't worry about subbing/unsubbing to an array of channels
    SUBSCRIBE { channel: &'a str },
    UNSUBSCRIBE { channel: &'a str },
    PUBLISH { channel: &'a str, msg: &'a str},
    PSUBSCRIBE { pattern: &'a str },
    PUNSUBSCRIBE { pattern: &'a str},
}

pub fn parse_message<'a> (msg: &'a str) -> Result<PubSubMessage<'a>, String> {
    let msg_contents: Vec<_> = msg.split(" ").collect();

    return match msg_contents.as_slice() {
        ["SUBSCRBE", channel] => Ok(PubSubMessage::SUBSCRIBE { channel }),
        ["UNSUBSCRIBE", channel] => Ok(PubSubMessage::UNSUBSCRIBE { channel }),
        ["PUBLISH", channel, msg] => Ok(PubSubMessage::PUBLISH { channel, msg }),
        _ => Err("Could not parse ws message.".to_string())
    }
}

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