我想要一个带有两个参数的函数,这两个参数都可以变成 的迭代器Foo
。问题是我愿意接受既是又是的IntoIterator<Foo>
东西IntoIterator<&Foo>
。重要的Foo
是Copy
,我可以从它的参考中廉价地创建一个拥有的副本。
我目前的解决方案是:
use std::borrow::Cow;
use std::iter::IntoIterator;
fn main() {
let foos = [Foo {}, Foo {}, Foo {}];
let references = || foos.iter();
let owned = || foos.iter().cloned();
bar(references(), references());
bar(references(), owned());
bar(owned(), references());
bar(owned(), owned());
}
#[derive(Copy, Clone)]
struct Foo {
// code ommitted here
}
impl<'a> From<Foo> for Cow<'a, Foo> {
fn from(foo: Foo) -> Self {
Self::Owned(foo)
}
}
impl<'a> From<&'a Foo> for Cow<'a, Foo> {
fn from(foo: &'a Foo) -> Self {
Self::Borrowed(foo)
}
}
fn bar<'a, AIter, A, BIter, B>(alpha_iter: AIter, beta_iter: BIter)
where
AIter: IntoIterator<Item=A>,
A: Into<Cow<'a, Foo>>,
BIter: IntoIterator<Item=B>,
B: Into<Cow<'a, Foo>>
{
for (alpha, beta) in alpha_iter.into_iter().zip(beta_iter.into_iter()) {
some_foo_specific_thing(*alpha.into(), *beta.into());
}
}
fn some_foo_specific_thing(alpha: Foo, beta: Foo) {
// code ommitted here
}
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有没有办法以更少的行数/无需执行此操作Cow
?
首先,你不需要IntoIterator
在这里精确绑定。这对于 来说就足够了Iterator<Item = Foo>
。
use std::borrow::Cow;
use std::iter::IntoIterator;
fn main() {
let foos = [Foo {}, Foo {}, Foo {}];
let references = || foos.iter(); // it is iterator itself
let owned = || foos.iter().cloned(); // it is iterator itself
bar(references(), references());
bar(references(), owned());
bar(owned(), references());
bar(owned(), owned());
}
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fn bar<'a, AIter, A, BIter, B>(alpha_iter: AIter, beta_iter: BIter)
where
AIter: Iterator<Item = A>,
A: Into<Cow<'a, Foo>>,
BIter: Iterator<Item = B>,
B: Into<Cow<'a, Foo>>
{
for (alpha, beta) in alpha_iter.zip(beta_iter) {
some_foo_specific_thing(*alpha.into(), *beta.into());
}
}
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其次,我认为最好使用Borrow
trait:
fn bar<'a, AIter, A, BIter, B>(alpha_iter: AIter, beta_iter: BIter)
where
AIter: Iterator<Item = A>,
A: std::borrow::Borrow<Foo> + Copy,
BIter: Iterator<Item = B>,
B: std::borrow::Borrow<Foo> + Copy,
{
let alpha_iter = alpha_iter.map(|item| *item.borrow()); // here we get a reference to an item and then make a copy to own it
let beta_iter = beta_iter.map(|item| *item.borrow()); // here we get a reference to an item and then make a copy to own it
for (alpha, beta) in alpha_iter.zip(beta_iter) {
some_foo_specific_thing(alpha, beta);
}
}
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