What does "expected struct `ParseError`, found fn item" mean?-CodePudding

I am working on a parser and I keep getting an error on this part of my code.

impl FromStr for Binop {
    type Err = ParseError;
    
    fn from_str(s: &str) -> Result<Self, Self::Err> {
        Ok(match s {
            " "     => Binop::Add,
            "*"     => Binop::Mul,
            "-"     => Binop::Sub,
            "/"     => Binop::Div,
            "<"     => Binop::Lt,
            "=="    => Binop::Eq,
            _   => {return Err(ParseError); } // <---- Error Here
        })
    }
}

I've tried writing an actual string inside the parenthesis and a bunch of other stuff & I just can't seem to understand what the error means.

Full error:

error[E0308]: mismatched types
   --> grumpy/src/isa.rs:212:32
    |
212 |             _   => {return Err(ParseError); }
    |                                ^^^^^^^^^^ expected struct `ParseError`, found fn item
    |
   ::: grumpy/src/lib.rs:17:1
    |
17  | pub struct ParseError(String);
    | ------------------------------ fn(String) -> ParseError {ParseError} defined here
    |
    = note: expected struct `ParseError`
              found fn item `fn(String) -> ParseError {ParseError}`
help: use parentheses to instantiate this tuple struct
    |
212 |             _   => {return Err(ParseError(_)); }
    |

And I defined the ParseError as pub struct ParseError(String);

CodePudding user response：

The named tuple requires a String value to be constructed. Note that a string literal has type &str; you need to invoke the to_string() method to turn it into a string:

Err(ParseError("foo".to_string()))

The compile error refers to fn item because by itself, in this context, ParseError refers to the implicit constructor function, which takes a String and returns a ParseError.