Can you help me understand? - day1 / part2 about advent-of-code

[BurntSushi]

Hi! I'm happy to help, but please do make sure to include all pertinent information. This should include the full code. I added the necessary imports and constant to your code for you this time. :-)

use std::collections::HashSet;
use std::fs;

const FILE_NAME: &'static str = "input.txt";

fn main() {
    let content = fs::read_to_string(FILE_NAME)
        .expect("Error while reading the input file");

    let mut previous = HashSet::new();
    let mut count = 0;

    previous.insert(count);

    for line in content.lines() {
        if !line.is_empty() {
            let current: i32 = line.parse()
                .expect("Error while parsing line");

            count += current;

            if previous.contains(&count) { // I NEVER PASS THIS CONDITION, WHY? :(
                println!("here! {}", count);
                break;
            }

            previous.insert(count);
        }
    }

    println!("{}", count);
}

Once doing that, and after dropping day 1's input into the current directory, running it appears to get part 1 right:

$ cargo run --release
   Compiling aoc-issue-3 v0.1.0 (/tmp/aoc-issue-3)
    Finished release [optimized] target(s) in 0.46s
     Running `target/release/aoc-issue-3`
437

For part 2, I think you might have misunderstood the puzzle. You need to cycle over the input as many times as is necessary to find a repetition. One loop through the input isn't enough. e.g.,

use std::collections::HashSet;
use std::fs;

const FILE_NAME: &'static str = "input.txt";

fn main() {
    let content = fs::read_to_string(FILE_NAME)
        .expect("Error while reading the input file");

    let mut previous = HashSet::new();
    let mut count = 0;

    previous.insert(count);

    loop {
        for line in content.lines() {
            if !line.is_empty() {
                let current: i32 = line.parse()
                    .expect("Error while parsing line");

                count += current;

                if previous.contains(&count) { // I NEVER PASS THIS CONDITION, WHY? :(
                    println!("here! {}", count);
                    return;
                }

                previous.insert(count);
            }
        }
    }
}

And here's the output:

$ cargo run --release
   Compiling aoc-issue-3 v0.1.0 (/tmp/aoc-issue-3)
    Finished release [optimized] target(s) in 0.43s
     Running `target/release/aoc-issue-3`
here! 655

Good luck!

[fabienjuif]

Thank you @BurntSushi this is much appreciated 👍

You are right, I'll take care of putting the whole code next time.
And yeah, this is shame that I didn't understand the puzzle right :(

Note that your device might need to repeat its list of frequency changes many times before a duplicate frequency is found, and that duplicates might be found while in the middle of processing the list.

Thanks!