我可以写一个 Rust for
循环等同于这个 C 代码吗:
for(int i = 2; i <= 128; i=i*i){
//do something
}
我只看到类似的东西
for i in 0..128 { /* do something */ }
或
let v = vec![0, 1, 2, /* ... */ ];
for i in v.iter() { /* do something */ }
我应该只使用 while
循环吗?
最佳答案
您可以始终创建一个自定义迭代器来执行您需要的任何唯一序列:
struct Doubling {
current: u64,
max: u64,
}
impl Iterator for Doubling {
type Item = u64;
fn next(&mut self) -> Option<Self::Item> {
if self.current > self.max {
None
} else {
let v = Some(self.current);
self.current *= 2;
v
}
}
}
fn main() {
let iter = Doubling { current: 2, max: 128 };
let values: Vec<_> = iter.collect();
println!("{:?}", values);
}
重要的是要认识到,当值加倍超出类型的大小时,这种逻辑(就像原来的 C!)有讨厌的边缘情况。
在这种特殊情况下,您还可以认识到您有一个指数级数:
fn main() {
let iter = (1..8).map(|p| 2i32.pow(p));
let values: Vec<_> = iter.collect();
println!("{:?}", values);
}
如果您想真正体验一下,请查看 Lazy sequence generation in Rust .改编于此:
#![feature(generators, generator_trait, conservative_impl_trait)]
use std::ops::{Generator, GeneratorState};
fn doubling(mut start: u64, max: u64) -> impl Iterator<Item = u64> {
GeneratorIteratorAdapter(move || {
while start <= max {
yield start;
start *= 2;
}
})
}
fn main() {
let iter = doubling(2, 128);
let sum: Vec<_> = iter.collect();
println!("{:?}", sum);
}
/* copy-pasta */
struct GeneratorIteratorAdapter<G>(G);
impl<G> Iterator for GeneratorIteratorAdapter<G>
where
G: Generator<Return = ()>,
{
type Item = G::Yield;
fn next(&mut self) -> Option<Self::Item> {
match self.0.resume() {
GeneratorState::Yielded(x) => Some(x),
GeneratorState::Complete(_) => None,
}
}
}
关于for-loop - 有没有办法做一个既不是迭代也不是线性的for循环?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/49030578/