CODEWITHSUNDEEP
rust
skatori
skatori

Reputation: 597

Programming in Rust , how to fix error[E0515] "cannot return value referencing local variable"?

Please help me to compile my code attached bellow. The compiler says that following 2 patterns depending on which lines I comment out.

The program reads a &str which is a simple "svg path command" like code then parses it. The pasted code has been simplified for simplicity. It uses Regex to split the input string into lines then study each line in the main for loop. Each loop pushes the parse result onto a vector. Finally the function returns the vector.

Basically the compiler says returning the vector is not allowed because it refers local variable. Though I don't have any workaround.

error[E0597]: `cmd` does not live long enough
  --> src/main.rs:24:25
   |
24 |    codeV =  re.captures(cmd.as_str());
   |    -----                ^^^ borrowed value does not live long enough
   |    |
   |    borrow might be used here, when `codeV` is dropped and runs the destructor for type `Option<regex::Captures<'_>>`
...
30 |    }
   |    - `cmd` dropped here while still borrowed
   |
   = note: values in a scope are dropped in the opposite order they are defined


error[E0515]: cannot return value referencing local variable `cmd`
  --> src/main.rs:31:1
   |
24 |    codeV =  re.captures(cmd.as_str());
   |                         --- `cmd` is borrowed here
...
31 | V //Error
   | ^ returns a value referencing data owned by the current function

Playground

use regex::Regex;
pub fn parse(path:&str) {//->Vec<Option<regex::Captures<>>>  //Error
    let reg_n=Regex::new(r"\n").unwrap();
    let path=reg_n.replace_all("\n"," ");
    let reg_cmd=Regex::new(r"(?P<cmd>[mlhv])").unwrap();
    let path=reg_cmd.replace_all(&path,"\n${cmd}");
    let cmdV=reg_n.split(&path);
   
    //let cmdV:Vec<&str> = reg.split(path).map(|x|x).collect();    

    let mut V:Vec<Option<regex::Captures<>>>=vec![];
    let mut codeV:Option<regex::Captures<>>=None;
    let mut count=0;
    for cmd_f in cmdV{//This loop block has been simplified.
        count+=1;
        if count==1{continue;}
        let mut cmd="".to_string();
   
        cmd=cmd_f.to_string();
        cmd=cmd.replace(" ","");
        let re = Regex::new(r"\{(?P<code>[^\{^\}]{0,})\}").unwrap();
        codeV =  re.captures(cmd.as_str());
        //cmd= re.replace_all(cmd.as_str(),"").to_string();
        let cmd_0=cmd.chars().nth(0).unwrap();
        //cmd.remove(0);
        //V.push(codeV);   //Compile error
        V.push(None); //OK
    }
    //V
}

fn main() {
    parse("m {abcd} l {efgh}");
}

Upvotes: 1

Views: 3321

Answers (2)

张国凯
张国凯

Reputation: 9

you can use cmd.clone().as_str() if you not sure the value has borrowed or no

Upvotes: -2

Masklinn
Masklinn

Reputation: 42282

Though I don't have any workaround.

regex's captures refer to the string they matched for efficiency. This means they can't outlive that string, as the match groups are essentially just offsets into that string.

Since the strings you match are created in the loop body, this means captures can't escape the loop body.

Aside from not creating strings in the loop body (or even the function), the solution / workaround is to convert your capture groups to owned data and store that: instead of trying to return a vector of captures, extract from the capture the data you actually want, convert it to an owned String (or tuple thereof, or whatever), and push that onto your vector.

e.g. https://play.rust-lang.org/?version=stable&mode=debug&edition=2018&gist=0107333e30f831a418d75b280e9e2f31

Upvotes: 3

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