如何交换两个变量? [英] How to swap two variables?
问题描述
与此 Python 代码最接近的等效 Rust 代码是什么?
What is the closest equivalent Rust code to this Python code?
a, b = 1, 2
a, b = b, a + b
我正在尝试编写一个迭代斐波那契函数.我有要转换为 Rust 的 Python 代码.一切都很好，除了交换部分.
I am trying to write an iterative Fibonacci function. I have Python code I want to convert to Rust. Everything is fine, except for the swap part.
def fibonacci(n):
if n < 2:
return n
fibPrev = 1
fib = 1
for num in range(2, n):
fibPrev, fib = fib, fib + fibPrev
return fib
推荐答案
在交换变量时，您最可能想要的是为 a 和 <创建新的绑定代码>b.
When swapping variables, the most likely thing you want is to create new bindings for a and b.
fn main() {
let (a, b) = (1, 2);
let (b, a) = (a, a + b);
}
但是，在您的实际情况中，稳定的 Rust 没有很好的解决方案.执行上述操作时，您总是为 a 和 b 创建新绑定，但您的案例想要修改现有绑定.我知道的一种解决方案是使用临时:
However, in your actual case, there isn't a nice solution in stable Rust. When you do as above, you always create new bindings for a and b, but your case wants to modify the existing bindings. One solution I know of is to use a temporary:
fn fibonacci(n: u64) -> u64 {
if n < 2 {
return n;
}
let mut fib_prev = 1;
let mut fib = 1;
for _ in 2..n {
let next = fib + fib_prev;
fib_prev = fib;
fib = next;
}
fib
}
你也可以让你改变元组:
You could also make it so that you mutate the tuple:
fn fibonacci(n: u64) -> u64 {
if n < 2 {
return n;
}
let mut fib = (1, 1);
for _ in 2..n {
fib = (fib.1, fib.0 + fib.1);
}
fib.1
}
在 nightly Rust 中，您可以使用 解构赋值:
In nightly Rust, you can use destructuring assignment:
#![feature(destructuring_assignment)]
fn fibonacci(n: u64) -> u64 {
if n < 2 {
return n;
}
let mut fib_prev = 1;
let mut fib = 1;
for _ in 2..n {
(fib_prev, fib) = (fib, fib + fib_prev);
}
fib
}
您可能还对交换两块内存的内容感兴趣.在 99+% 的情况下，您希望重新绑定变量，但在极少数情况下您希望就地"更改内容:
You may also be interested in swapping the contents of two pieces of memory. 99+% of the time, you want to re-bind the variables, but a very small amount of time you want to change things "in place":
fn main() {
let (mut a, mut b) = (1, 2);
std::mem::swap(&mut a, &mut b);
println!("{:?}", (a, b));
}
请注意，在一个步骤中进行这种交换并将值加在一起并不简洁.
Note that it's not concise to do this swap and add the values together in one step.
对于返回迭代器的斐波那契数列的非常简洁的实现:
For a very concise implementation of the Fibonacci sequence that returns an iterator:
fn fib() -> impl Iterator<Item = u128> {
let mut state = [1, 1];
std::iter::from_fn(move || {
state.swap(0, 1);
let next = state.iter().sum();
Some(std::mem::replace(&mut state[1], next))
})
}
另见:
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