iterator - 使用推断的(？)类型克隆std::iter::Map

我很难用紧凑的方式克隆地图：

extern crate itertools_num;

use itertools_num::linspace;

fn main() {
    // 440Hz as wave frequency (middle A)
    let freq: f64 = 440.0;
    // Time vector sampled at 880 times/s (~Nyquist), over 1s
    let delta: f64 = 1.0 / freq / 2.0;
    let time_1s = linspace(0.0, 1.0, (freq / 2.0) as usize)
        .map(|sample| { sample * delta});

    let sine_440: Vec<f64> = time_1s.map(|time_sample| {
        (freq * time_sample).sin()
    }).collect();

    let sine_100: Vec<f64> = time_1s.map(|time_sample| {
        (100.0 * time_sample).sin()
    }).collect();
}

这个代码的错误是

`time_1s` moved here because it has type `std::iter::Map<itertools_num::Linspace<f64>, [closure@examples/linear_dft.rs:12:14: 12:40 delta:&f64]>`, which is non-copyable

这是可以理解的，但是如果我尝试使用time_1s.clone()来代替

note: the method `clone` exists but the following trait bounds were not satisfied: `[closure@examples/linear_dft.rs:12:14: 12:40 delta:_] : std::clone::Clone`
error: the type of this value must be known in this context
     (freq * time_sample).sin()

这也是可以理解的，但是在返回之前将(freq * time_sample).sin()存储在闭包内的let foo: f64中没有任何效果。
在这种情况下我该怎么办？我只想不止一次地使用时间向量。

最佳答案

使用time_1s两次的一种方法是同时使用并在最后解压缩：

extern crate itertools_num;

use itertools_num::linspace;

fn main() {
    // 440Hz as wave frequency (middle A)
    let freq: f64 = 440.0;
    // Time vector sampled at 880 times/s (~Nyquist), over 1s
    let delta: f64 = 1.0 / freq / 2.0;
    let time_1s = linspace(0.0, 1.0, (freq / 2.0) as usize)
            .map(|sample| { sample * delta});

    let (sine_440, sine_100): (Vec<f64>, Vec<f64>) = time_1s.map(|time_sample| {
        ((freq * time_sample).sin(),
         (100.0 * time_sample).sin())
    }).unzip();
}