试图对 Vec() 的成员进行可变借用，但 rustc 认为我正在尝试可变借用 Vec()

Question

In the assembler I'm working on, identifiers can be of two types - labels or symbols. When being defined, the tokenize_line() helper function can automatically set the identifier type argument of the Identifier variant of the TokenInfo enum to Some(IdentifierType::Label) or Some(IdentifierType::Symbol). However, when an identifier is being used the identifier type is not immediately obvious as it is when defining identifiers.

As such, this needs to be done in the main lex() function. However, in order to do this, I need to grab mutable borrows of all the identifier tokens because unless the identifier type is already Some(whatever) I will have to modify them. Part of this was doing a linear search through my vector of vectors of tokens (list of list of tokens) and grabbing the positions of all identifier tokens and the second part is actually making mutable borrows of all those identifier tokens and putting those into a vector/list.

However, doing this through a for loop creates a problem: the rust compiler thinks I'm repeatedly borrowing the vector/list, and not the individual tokens. I have no idea if this can be worked around, or if I'll just have to use the position list and the actual list of lists of tokens. Here's my code:

let mut i = 0; // i = Line counter
let mut j = 0; // j = Token counter
let mut positions: Vec<(usize, usize)> = vec!();
for line in &tokens {
    for token in line {
        match &token.info {
            TokenInfo::Identifier(t) => {
                positions.push((i, j));
            }
            _ => {}
        }
        j = j + 1;
    }
    i = i + 1;
}

let mut ident_tokens: Vec<&mut Token> = vec!();

for position in positions {
    ident_tokens.push(&mut tokens[position.0][position.1]);
}

Answer 1

The reference returned by the [] syntax is considered to be derived from the reference to the container, so you can't have two such references at the same time. Therefore, instead of

fn vec_of_ref_mut(vec: &mut Vec<i32>) -> Vec<&mut i32> {
    let mut result = vec![];
    
    for i in 0..vec.len() {
        result.push(&mut vec[i]);
    }
    
    result
}

you would need to use iter_mut :

fn vec_of_ref_mut(vec: &mut Vec<i32>) -> Vec<&mut i32> {
    vec.iter_mut().collect()
}

In a nested case like yours, you would need to use flat_map . For example:

fn vec_of_ref_mut(vec: &mut Vec<Vec<i32>>) -> Vec<&mut i32> {
    vec.iter_mut().flat_map(|sub_vec| sub_vec.iter_mut()).collect()
}