我有一个通用的结构Dit<T>
,用于实现T的FFT:
struct Dit<T> {
n: usize,
exponents: Bin<f32, Complex<T>>,
tmp: Option<Vec<Complex<T>>>,
}
impl Dit<f32> {
/// Create a new instance
///
/// Notice that the number of samples that will be processed by an instance
/// is always fixed, because the exponent values are precalculated.
///
/// # Parameters
/// - `n` The number of samples this operator can process, eg. 1024
pub fn new(n: usize) -> Result<Dit<f32>, FFTError> {
if 2.pow((n as f64).log2() as usize) != n {
return Err(FFTError::InvalidLength);
}
let rtn = Dit {
n: n,
exponents: Bin::new(),
tmp: None,
}.pregen();
return Ok(rtn);
}
// ...
}
我开始添加f64
的实现:
impl Dit<f64> {
pub fn new(n: usize) -> Result<Dit<f64>, FFTError> {
unimplemented!()
}
// ...
}
......我遇到了这些错误:
src/impls/dit.rs:186:7: 196:4 error: duplicate definition of value `new`
src/impls/dit.rs:186 pub fn new(n:usize) -> Result<Dit<f64>, FFTError> {
src/impls/dit.rs:187 if 2.pow((n as f64).log2() as usize) != n {
src/impls/dit.rs:188 return Err(FFTError::InvalidLength);
src/impls/dit.rs:189 }
src/impls/dit.rs:190 let rtn = Dit {
src/impls/dit.rs:191 n: n,
...
src/impls/dit.rs:110:7: 120:4 note: first definition of value `new` here
src/impls/dit.rs:110 pub fn new(n:usize) -> Result<Dit<f32>, FFTError> {
src/impls/dit.rs:111 if 2.pow((n as f64).log2() as usize) != n {
src/impls/dit.rs:112 return Err(FFTError::InvalidLength);
src/impls/dit.rs:113 }
src/impls/dit.rs:114 let rtn = Dit {
src/impls/dit.rs:115 n: n,
我很困惑。我的印象是,对于通用Foo<T>
,实现Foo<Bar1>
是Foo<Bar2>
实现的不同具体实例。因此,我的印象是我可以为每个具体实例提供不同的方法实例。
我做错了什么?
答案 0 :(得分:3)
我认为用这种语法解决你的任务是不可能的(至少,我在Rust参考书中找不到任何例子)。
但是有一些工作结构如:
impl<T> Dit<T> where T: Float {
或:
trait DitTrait {
fn new(n: usize) -> Result<Self, FFTError>;
}
impl DitTrait for Dit<f32> { ... }
impl DitTrait for Dit<f64> { ... }