How do I clone a closure, so that their types are the same?

Question

I have a struct which looks something like this:

pub struct MyStruct
where
    F: Fn(usize) -> f64,
{
    field: usize,
    mapper: F,
    // fie         


        

Answer 1

You can use trait objects to be able to implement Сlone for your struct:

use std::rc::Rc;

#[derive(Clone)]
pub struct MyStructRef<'f>  {
    field: usize,
    mapper: &'f Fn(usize) -> f64,
}


#[derive(Clone)]
pub struct MyStructRc  {
    field: usize,
    mapper: Rc f64>,
}

fn main() {
    //ref
    let closure = |x| x as f64;
    let f = MyStructRef { field: 34, mapper: &closure };
    let g = f.clone();
    println!("{}", (f.mapper)(3));
    println!("{}", (g.mapper)(3));

    //Rc
    let rcf = MyStructRc { field: 34, mapper: Rc::new(|x| x as f64 * 2.0) };
    let rcg = rcf.clone();
    println!("{}", (rcf.mapper)(3));
    println!("{}", (rcg.mapper)(3));    
}