How to fix this problem?

help
1

Want to use for loop instead of a long chain of .remove(b' ') fn calls for percent_encoding::NON_ALPHANUMERIC

like:

percent_encoding::NON_ALPHANUMERIC.remove(b'.')
.remove(b':')
.remove(b'%')
.remove(..)...

convert to:

let r = cell::RefCell::new(percent_encoding::NON_ALPHANUMERIC);

for c in b".:/-_?=%" {
    let mut bm = r.borrow_mut();
    *bm = &bm.remove(*c);
}

got consider using a let binding to create a longer lived value error.

Any suggestions?

2

Computing the ASCII set at runtime seems unnecessary to me. I.e. what is the benefit using a for loop over creating your own constant by chaining .remove calls on NON_ALPHANUMERIC?

3

if want to .remove() 10 or 20 more chars NON_ALPHANUMERIC.remove(b' ').remove().remove()... 20 times long chain is hard to write and read. Is there a way to change it at runtime using for loop?

4

I disagree. But you can do something like this to compute your AsciiSet at both compile and run time using a while loop:

use percent_encoding::{AsciiSet, NON_ALPHANUMERIC};

pub const fn my_ascii_set() -> AsciiSet {
    const REM: &[u8] = b".:/-_?=%";
    
    // We need to call `remove` here once to get an owned instance of `AsciiSet` 
    // from `NON_ALPHANUMERIC`, which is a `&AsciiSet`.
    let mut set = NON_ALPHANUMERIC.remove(REM[0]);
    
    let mut i = 1;
    
    while i < REM.len() {
        set = set.remove(REM[i]);
        i += 1;
    }
    
    set
}

Playground.

If you really don't want to do this at compile time, you can substitute the while loop with a for loop:

use percent_encoding::{AsciiSet, NON_ALPHANUMERIC};

pub fn my_ascii_set() -> AsciiSet {
    const REM: &[u8] = b".:/-_?=%";
    
    // We need to call `remove` here once to get an owned instance of `AsciiSet` 
    // from `NON_ALPHANUMERIC`, which is a `&AsciiSet`.
    let mut set = NON_ALPHANUMERIC.remove(REM[0]);
    
    for c in &REM[1..] {
        set = set.remove(*c);
    }
    
    set
}

Playground.