How to fold using a HashMap as an accumulator? - rust

Question

This code works:
let stdin = std::io::stdin();
let mut rdr = csv::Reader::from_reader(stdin);
let mut hmap = HashMap::<String, u64>::new();
rdr.records()
.map(|r| r.unwrap())
.fold((), |_, item| {
// TODO: Is there a way not to have to copy item[col] every time?
let counter = hmap.entry(item[col].to_string()).or_insert(0);
*counter += 1;
});
This code fails with the message: "cannot move out of acc because it is borrowed"
let stdin = std::io::stdin();
let mut rdr = csv::Reader::from_reader(stdin);
let hmap = rdr.records()
.map(|r| r.unwrap())
.fold(HashMap::<String, u64>::new(), |mut acc, item| {
// TODO: Is there a way not to have to copy item[col] every time?
let counter = acc.entry(item[col].to_string()).or_insert(0);
*counter += 1;
acc
});

Answer 1

You cannot return acc from the closure because you have a mutable borrow to it that still exists (counter).
This is a limitation of the Rust compiler (specifically the borrow checker). When non-lexical lifetimes are enabled, your original code will work:
#![feature(nll)]
use std::collections::HashMap;
fn main() {
let hmap = vec![1, 2, 3].iter().fold(HashMap::new(), |mut acc, _| {
let counter = acc.entry("foo".to_string()).or_insert(0);
*counter += 1;
acc
});
println!("{:?}", hmap);
}
Before NLL, the compiler is overly conservative about how long a borrow will last. To work around this, you can introduce a new scope to constrain the mutable borrow:
use std::collections::HashMap;
fn main() {
let hmap = vec![1, 2, 3].iter().fold(HashMap::new(), |mut acc, _| {
{
let counter = acc.entry("foo".to_string()).or_insert(0);
*counter += 1;
}
acc
});
println!("{:?}", hmap);
}
You can also prevent the borrow from lasting beyond the line it's needed in:
use std::collections::HashMap;
fn main() {
let hmap = vec![1, 2, 3].iter().fold(HashMap::new(), |mut acc, _| {
*acc.entry("foo".to_string()).or_insert(0) += 1;
acc
});
println!("{:?}", hmap);
}
I assumed Rust would know that counter would go out of scope once acc was returned
This is understandable and relates to the non-lexical lifetimes discussion. The "good" news is that Rust is being consistent about how references work when the thing being referenced moves. In this case, you are moving the accumulator into an "output slot". You can see this with plain functions as well:
fn foo(mut s: Vec<u8>) -> Vec<u8> {
let borrow = &mut s[0];
s
}
fn main() {}
But really, it's the same as moving a referred-to variable at all:
fn main() {
let mut s = Vec::<u8>::new();
let borrow = &mut s[0];
let s2 = s;
}
Both of these fail before NLL and work afterwards.