如何理解下面这段代码？我是Rust的新手,但有C/Haskell背景和一点点C++.我能找到的唯一参考是减轻强制.
fn main() {
let xs: [u32; 4] = [0, 1, 2, 3];
let mut i: u32 = 0;
for x in xs.iter() {
if i > *x { // It looks x is an iterator. Understood.
i = i + x; // no error. (coerced)
//Quote: "Rust will do this as many times
// as possible until the types match."
i = i + *x; // no error (explicit deref)
i += x; // error about u32/&u32 …Consider this example:
fn main() {
let v: Vec<i32> = vec![1, 2, 3, 4, 5];
let b: i32 = (&v[2]) * 4.0;
println!("product of third value with 4 is {}", b);
}
This fails as expected as float can't be multiplied with &i32.
error[E0277]: cannot multiply `{float}` to `&i32`
--> src\main.rs:3:23
|
3 | let b: i32 = (&v[2]) * 4.0;
| ^ no implementation for `&i32 * {float}`
|
= help: the trait `std::ops::Mul<{float}>` is not implemented for …我有以下代码:
fn example(known_primes: &[i32], number: i32, prime: i32, limit: i32) {
let mut is_prime = true;
for prime in known_primes {
if number % prime == 0 {
is_prime = false;
break;
}
if *prime > limit {
break;
}
}
}
为什么我需要prime在第二个条件(*prime > limit)中取消引用,当我不需要在第一个条件()中这样做时number % prime == 0？
这两个%和<是两个数字并把返回一些运营商.唯一的区别似乎在于它们返回的内容(数字与布尔值).虽然为什么不能将借用的整数与文字整数进行比较？ 确实解释了使代码工作所需的内容(所有重载的实现,理想情况下在标准库中),它没有说明它为什么起作用a % b.这些运营商之间是否存在根本区别？或者它还没有实现？