rust program

fn main() {
    let args: Vec<String> = env::args().collect();

    let parse_args = Config::build(&args);

    dbg!(args);
    let config = parse_args.unwrap();
    dbg!(config.file_path);
    dbg!(config.query);
}

#[derive(Debug)]
struct Config {
    query: String,
    file_path: String
}

impl Config {
    fn new(args: &[String]) -> Config {
        return Config{
            query: args[1].clone(),
            file_path: args[2].clone()
        }
    }

    fn build(args: &[String]) -> Result<Config, &'static str> {
        if args.len() < 3 {
            return Err("error num");
        }
        return Ok(Self::new(args));
    }
}

but when i remove 'static in build method, compile error

fn build(args: &[String]) -> Result<Config, &str>

my question is, why i change &'static str to &str, i will compile fail

cargo check result

error[E0505]: cannot move out of `args` because it is borrowed  
4 |     let args: Vec<String> = env::args().collect();  
  |         ---- binding `args` declared here
5 |
6 |     let parse_args = Config::build(&args);
  |                                    ----- borrow of `args` occurs here
7 |
8 |     dbg!(args);
  |     ^^^^^^^^^^ move out of `args` occurs here
9 |     let config = parse_args.unwrap();
  |                  ---------- borrow later used here

it become a borrowed error

rust program

fn main() {
    let args: Vec<String> = env::args().collect();

    let parse_args = Config::build(&args);

    dbg!(args);
    let config = parse_args.unwrap();
    dbg!(config.file_path);
    dbg!(config.query);
}

#[derive(Debug)]
struct Config {
    query: String,
    file_path: String
}

impl Config {
    fn new(args: &[String]) -> Config {
        return Config{
            query: args[1].clone(),
            file_path: args[2].clone()
        }
    }

    fn build(args: &[String]) -> Result<Config, &'static str> {
        if args.len() < 3 {
            return Err("error num");
        }
        return Ok(Self::new(args));
    }
}

but when i remove 'static in build method, compile error

fn build(args: &[String]) -> Result<Config, &str>

my question is, why i change &'static str to &str, i will compile fail

cargo check result

error[E0505]: cannot move out of `args` because it is borrowed  
4 |     let args: Vec<String> = env::args().collect();  
  |         ---- binding `args` declared here
5 |
6 |     let parse_args = Config::build(&args);
  |                                    ----- borrow of `args` occurs here
7 |
8 |     dbg!(args);
  |     ^^^^^^^^^^ move out of `args` occurs here
9 |     let config = parse_args.unwrap();
  |                  ---------- borrow later used here

it become a borrowed error

• rust

• Please edit your question to include the full error message as reported by cargo check (not from your IDE). – Jmb Commented Apr 2 at 6:59

1 Answer 1

Because of the Rust lifetime elision rules, this code:

fn build(args: &[String]) -> Result<Config, &str>

is equivalent to:

fn build<'a>(args: &'a [String]) -> Result<Config, &'a str>

which links the lifetime of the Result to args and means that args stays borrowed until you unwrap the Result. But dbg!(args) tries to take ownership while the parse_args borrow is still alive.