Rust：从仅<T> 不同的函数返回通用结构

如何解决Rust：从仅<T> 不同的函数返回通用结构

我正在寻求有关正确语法或 Rust 方法的帮助。我的用例：我有一个通用结构 FileData，它有一个名为 provider 的变量。提供者必须实现 AsRef<[u8]> 以便数据可能来自静态字节、堆分配的内存、内存映射，以及其他可能的数据。我有几种创建 FileData 的方法，它们似乎运行良好。但是有一个

// ERROR: This is the line that I do not get right 
pub fn from_file<P: AsRef<Path>>(filename: P,mmap: bool) -> Result<FileData<T>,Box<dyn Error>> {
    if mmap == true {
        return FileData::mmap_file(filename)
    } else {
        return FileData::read_file(filename)
    }
}

我不明白。该方法总是返回一个 FileData，返回取决于 'mmap' 参数，<T> 是不同的。它可以是 <Box<[u8]> 或 <Mmap>。

我搜索了类似的问题和文章，但可以找到符合我情况的，例如(1)、(2)、(3)。

#[derive(Debug)]
pub struct FileData<T: AsRef<[u8]>> {
    pub filename: String,pub provider: T,// data block,file read,mmap,and potentially more
    pub fsize: u64,pub mmap: bool,}

impl FileData<&[u8]> {
    /// Useful for testing. Create a FileData builder based on some bytes. 
    #[allow(dead_code)]
    pub fn from_bytes(data: &'static [u8]) -> Self {
        FileData {
            filename: String::new(),provider: data,fsize: data.len() as _,mmap: false,}
    }
}

pub fn path_to_string<P: AsRef<Path>>(filename: P) -> String {
    return String::from(filename.as_ref().to_str().unwrap_or_default());
}

pub fn file_size(file: &File) -> Result<u64,Box<dyn Error>> {
    Ok(file.Metadata()?.len())
}

impl FileData<Box<[u8]>> {
    /// Read the full file content into memory,which will be allocated on the heap.
    #[allow(dead_code)]
    pub fn read_file<P: AsRef<Path>>(filename: P) -> Result<Self,Box<dyn Error>> {
        let mut file = File::open(&filename)?;
        let fsize = file_size(&file)?;

        let mut provider = vec![0_u8; fsize as usize].into_Boxed_slice();
        let n = file.read(&mut provider)? as u64;
        assert!(fsize == n,"Failed to read all data from file: {} vs {}",n,fsize);

        Ok(FileData {
            filename: path_to_string(&filename),provider: provider,fsize: fsize,})
    }
}

impl FileData<Mmap> {
    /// Memory Map the file content
    #[allow(dead_code)]
    pub fn mmap_file<P: AsRef<Path>>(filename: P) -> Result<Self,Box<dyn Error>> {
        let file = File::open(&filename)?;
        let fsize = file_size(&file)?;
        let provider = unsafe { MmapOptions::new().map(&file)? };

        Ok(FileData {
            filename: path_to_string(&filename),mmap: true,})
    }
}

impl<T: AsRef<[u8]>> FileData<T> {
    #[allow(dead_code)]
    pub fn from_file<P: AsRef<Path>>(filename: P,mmap: bool) -> Result<FileData<_>,Box<dyn Error>> {
        if mmap == true {
            return FileData::mmap_file(filename)
        } else {
            return FileData::read_file(filename)
        }
    }

    pub fn as_ref(&self) -> &[u8] {
        return self.provider.as_ref()
    }
}

错误信息是：

error[E0308]: mismatched types
  --> src\data_files\file_data.rs:87:20
   |
83 | impl<T: AsRef<[u8]>> FileData<T> {
   |      - this type parameter
84 |     #[allow(dead_code)]
85 |     pub fn from_file<P: AsRef<Path>>(filename: P,Box<dyn Error>> {
   |                                                                  ----------------------------------- expected `std::result::Result<file_data::FileData<T>,std::Boxed::Box<(dyn std::error::Error + 'static)>>` because of return type
86 |         if mmap == true {
87 |             return FileData::mmap_file(filename)
   |                    ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ expected type parameter `T`,found struct `Mmap`
   |
   = note: expected enum `std::result::Result<file_data::FileData<T>,_>`
              found enum `std::result::Result<file_data::FileData<Mmap>,_>`

解决方法

泛型让调用者有权决定函数的返回类型应该是什么。现在你的函数，被调用者，正在决定返回类型，这就是你收到编译器错误的原因。

您可以重构代码，通过实现额外的特征 IntoFileData，然后将其添加为绑定到您的通用 FileData<T> 实现的特征，将权利返还给调用者。简化注释示例：

use memmap::Mmap;
use memmap::MmapOptions;
use std::error::Error;
use std::fs::File;
use std::io::Read;
use std::path::Path;

// simplified FileData for brevity
struct FileData<T: AsRef<[u8]>> {
    provider: T,}

// new trait for converting types into FileData
trait IntoFileData<T: AsRef<[u8]>> {
    fn from_path(path: &Path) -> Result<FileData<T>,Box<dyn Error>>;
}

impl IntoFileData<Box<[u8]>> for Box<[u8]> {
    fn from_path(path: &Path) -> Result<FileData<Box<[u8]>>,Box<dyn Error>> {
        let mut file = File::open(path)?;
        let size = file.metadata()?.len();

        let mut provider = vec![0_u8; size as usize].into_boxed_slice();
        let read = file.read(&mut provider)? as u64;
        assert!(
            size == read,"Failed to read all data from file: {} vs {}",read,size
        );

        Ok(FileData { provider })
    }
}

impl IntoFileData<Mmap> for Mmap {
    fn from_path(path: &Path) -> Result<FileData<Mmap>,Box<dyn Error>> {
        let file = File::open(path)?;
        let provider = unsafe { MmapOptions::new().map(&file)? };

        Ok(FileData { provider })
    }
}

// this signature gives the caller the right to choose the type of FileData
impl<T: AsRef<[u8]> + IntoFileData<T>> FileData<T> {
    fn from_path(path: &Path) -> Result<FileData<T>,Box<dyn Error>> {
        T::from_path(path)
    }
}

fn example(path: &Path) {
    // caller asks for and gets file data as Box<[u8]>
    let file_data: FileData<Box<[u8]>> = FileData::from_path(path).unwrap();

    // caller asks for and gets file data as Mmap
    let file_data: FileData<Mmap> = FileData::from_path(path).unwrap();
}

playground

---

如果你想给被调用者决定返回类型的权利，你必须返回一个 trait 对象。简化注释示例：

use memmap::Mmap;
use memmap::MmapOptions;
use std::error::Error;
use std::fs::File;
use std::io::Read;
use std::path::Path;

// simplified FileData for brevity
struct FileData {
    provider: Box<dyn AsRef<[u8]>>,}

fn vec_from_path(path: &Path) -> Result<FileData,Box<dyn Error>> {
    let mut file = File::open(path)?;
    let size = file.metadata()?.len();

    let mut provider = vec![0_u8; size as usize];
    let read = file.read(&mut provider)? as u64;
    assert!(
        size == read,size
    );

    Ok(FileData {
        provider: Box::new(provider),})
}

fn mmap_from_path(path: &Path) -> Result<FileData,Box<dyn Error>> {
    let file = File::open(path)?;
    let provider = unsafe { MmapOptions::new().map(&file)? };

    Ok(FileData {
        provider: Box::new(provider),})
}

impl FileData {
    fn from_path(path: &Path,mmap: bool) -> Result<FileData,Box<dyn Error>> {
        if mmap {
            mmap_from_path(path)
        } else {
            vec_from_path(path)
        }
    }
}

fn example(path: &Path) {
    // file data could be vec or mmap,callee decides
    let file_data = FileData::from_path(path,true).unwrap();
    let file_data = FileData::from_path(path,false).unwrap();
}

playground