我在以紧凑的方式克隆地图时遇到问题:
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();
}
我用这段代码得到的错误是
`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,但如果我尝试使用time_1s.clone()
相反,我得到
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()
这也可以理解,但是存储(freq * time_sample).sin()
in a let foo: f64
返回之前在闭包内没有任何效果。
在这种情况下我该怎么办?我想做的就是多次使用时间向量。
一种使用方法time_1s
两次是一起执行并在最后解压:
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();
}
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