如果我想在稳定的Rust中使用自定义步骤进行迭代,我该怎么办?基本上就像C/C++
for (int i = 0; i < n; i += 2) {
}
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我已经尝试过使用range_step_inclusive和解决方案如何使用自定义步骤迭代范围?:
use std::iter::range_step_inclusive;
for i in range_step_inclusive(0, n, 2) {
println!("i: {}", i);
}
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但似乎它在Rust 1.1中不可用:
error: unresolved import `std::iter::range_step_inclusive`. There is no `range_step_inclusive` in `std::iter`
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什么是另类?可能是创建自定义范围的惯用方法.
Iterator::step_by 现在稳定了:
fn main() {
for i in (0..100).step_by(2) {
println!("{}", i);
}
}
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你总是可以用老式的方式写出来:
fn main() {
let mut i = 0;
while i < 100 {
println!("i: {}", i);
i += 2;
}
}
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然后可以抽象出来:
use std::ops::Add;
fn step_by<T, F>(start: T, end_exclusive: T, step: T, mut body: F)
where
T: Add<Output = T> + PartialOrd + Copy,
F: FnMut(T),
{
let mut i = start;
while i < end_exclusive {
body(i);
i = i + step;
}
}
fn main() {
step_by(0, 100, 2, |i| {
println!("i: {}", i);
})
}
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有趣的历史旁注,我相信在迭代器变得极为普遍之前,最初所有的循环都是用这样的闭包完成的.
然后你可以把它变成迭代器:
use std::ops::Add;
struct StepBy<T> {
start: T,
end_exclusive: T,
step: T,
}
impl<T> StepBy<T> {
fn new(start: T, end_exclusive: T, step: T) -> Self {
Self {
start,
end_exclusive,
step,
}
}
}
impl<T> Iterator for StepBy<T>
where
T: Add<Output = T> + PartialOrd + Copy,
{
type Item = T;
fn next(&mut self) -> Option<Self::Item> {
if self.start < self.end_exclusive {
let v = self.start;
self.start = self.start + self.step;
Some(v)
} else {
None
}
}
}
fn main() {
for i in StepBy::new(0, 100, 2) {
println!("i: {}", i);
}
}
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也可以看看:
有使用let"重新定义"的方法:
for i in 0..((n + 1) / 2) {
let i = i * 2;
// …
}
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或使用Iterator::map:
for i in (0..((n + 1) / 2)).map(|i| i * 2) {
// …
}
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