如何将类型的“缺失泛型”指定为函数参数？

Question

我试图使用一个共享的信号量使用futures_intrusive::sync::GenericSemaphore

use std::path::Path;
use futures_intrusive::sync::GenericSemaphore;

async fn refresh_file(raster_file: &Path, slr: f64, output_path: &Path, chunk_size: u32, semaphore: &GenericSemaphore) {
    if output_path.exists().not() {
        build_required_windows(&raster_file, slr, chunk_size, output_path, &semaphore).await?;
    }
}

然而，我得到了一个错误：

error[E0107]: missing generics for struct `GenericSemaphore`
   --> src/linear_bathtubbing.rs:129:106
    |
129 | pub async fn refresh_file(raster_file: &Path, slr: f64, output_path: &Path, chunk_size: u32, semaphore: &GenericSemaphore) {
    |                                                                                                          ^^^^^^^^^^^^^^^^ expected 1 generic argument
    |
note: struct defined here, with 1 generic parameter: `MutexType`
   --> /.../.cargo/registry/src/github.com-1ecc6299db9ec823/futures-intrusive-0.4.0/src/sync/semaphore.rs:433:12
    |
433 | pub struct GenericSemaphore<MutexType: RawMutex> {
    |            ^^^^^^^^^^^^^^^^ ---------
help: add missing generic argument
    |
129 | pub async fn refresh_file(raster_file: &Path, slr: f64, output_path: &Path, chunk_size: u32, semaphore: &GenericSemaphore<MutexType>) {
    |                                                                                                          ~~~~~~~~~~~~~~~~~~~~~~~~~~~

似乎我需要“添加缺少的泛型参数”，但我不知道如何继续。我尝试过指定Mutex类型和其他东西，但没有成功，但我认为这可能有一个简单的答案。

Answer 1

只需将GenericSemaphore期望的泛型类型参数添加到函数签名中即可。

async fn refresh_file<T: RawMutex>(
    raster_file: &Path,
    slr: f64,
    output_path: &Path,
    chunk_size: u32,
    semaphore: &GenericSemaphore<T>
) {

或者，使用impl语法(对于上面的代码来说，这是糖)：

async fn refresh_file(
    raster_file: &Path,
    slr: f64,
    output_path: &Path,
    chunk_size: u32,
    semaphore: &GenericSemaphore<impl RawMutex>
) {