Fix lints

Author: maneatingape

src/year2016/day04.rs

@@ -23,7 +23,7 @@ pub fn parse(input: &str) -> Vec<Room<'_>> {
         let size = line.len();
         let name = &line[..size - 11];
         let sector_id = (&line[size - 10..size - 7]).unsigned();
-        let checksum = line[size - 6..size - 1].as_bytes();
+        let checksum = &line.as_bytes()[size - 6..size - 1];
 
         // Count the frequency of each digit, the frequency of each frequency `fof` and the
         // highest total frequency.

src/year2018/day12.rs

@@ -6,7 +6,7 @@
 //! The trick for part two is that the plants will eventually stabilize into a repeating pattern
 //! that expands by the same amount each generation. Once 2 deltas between 3 subsequent
 //! generations are the same we extrapolate 50 billion generations into the future.
-use std::iter::repeat;
+use std::iter::repeat_n;
 
 pub struct Input {
     rules: Vec<usize>,
@@ -73,7 +73,7 @@ fn step(rules: &[usize], tunnel: &Tunnel) -> Tunnel {
     let mut plants = Vec::with_capacity(1_000);
 
     // Add four extra empty pots to the end to make checking the last pattern easier.
-    for plant in tunnel.plants.iter().chain(repeat(&0).take(4)) {
+    for plant in tunnel.plants.iter().copied().chain(repeat_n(0, 4)) {
         index = ((index << 1) | plant) & 0b11111;
 
         sum += position * rules[index] as i64;

src/year2019/day12.rs

@@ -23,9 +23,11 @@ type Input = [Axis; 3];
 /// Group each axis together
 pub fn parse(input: &str) -> Input {
     let n: Vec<_> = input.iter_signed().collect();
-    [[n[0], n[3], n[6], n[9], 0, 0, 0, 0], [n[1], n[4], n[7], n[10], 0, 0, 0, 0], [
-        n[2], n[5], n[8], n[11], 0, 0, 0, 0,
-    ]]
+    [
+        [n[0], n[3], n[6], n[9], 0, 0, 0, 0],
+        [n[1], n[4], n[7], n[10], 0, 0, 0, 0],
+        [n[2], n[5], n[8], n[11], 0, 0, 0, 0],
+    ]
 }
 
 pub fn part1(input: &Input) -> i32 {

src/year2022/day16.rs

@@ -64,7 +64,7 @@ pub struct Input {
 /// State of a single exploration path through the valves.
 ///
 /// * `todo` Binary mask of unopened valves. For example if there are 3 unopened valves left this
-///    could look like `11100`.
+///   could look like `11100`.
 /// * `from` Index of current valve.
 /// * `time` The *remaining* time left.
 /// * `pressure` Total pressure released from all opened valves including future extrapolated flow.

src/year2023/day22.rs

@@ -56,7 +56,7 @@
 //! * `B` and `F` are both supported only by `A` so their depth is 1.
 //! * `C` will fall if either `A` or `B` is removed so its depth is 2.
 //! * `D` will only fall when `A` is removed. Removing `F` would leave it supported by `B` and `C`
-//!    or vice-versa. The common ancestor of the path to the root is `A` so its depth is 1.
+//!   or vice-versa. The common ancestor of the path to the root is `A` so its depth is 1.
 //! * `E`'s common ancestor is the floor so its depth is 0.
 //!
 //! In total `0 (A) + 0 (G) + 1 (B) + 1 (F) + 2 (C) + 1 (D) + 0 (E) = 5` bricks will fall.