难以理解 Rust 中的特征和泛型

Question

我对尝试编写泛型非常陌生，而且 Rust 的特性一直让我难以理解。我有这段代码：

pub trait Mapper {
    fn prg_rom_read(&self, addr: u16) -> u8 {}
    fn prg_rom_write(&mut self, addr: u16, val: u8) {}
    fn chr_rom_read(&self, addr: u16) -> u8 {}
    fn chr_rom_write(&mut self, addr: u16, val: u8) {}
}

pub fn choose_mapper<M: Mapper>(rom_header: &RomHeader) -> M {
    match rom_header.mapper_number {
        0 => Mapper1::new(rom_header),
        _ => panic!("Unsupported mapper: {:#}", rom_header.mapper_number),
    }
}

struct Mapper1 {
    prg_ram: Box<[u8]>,
    prg_rom: Box<[u8]>,
    chr: Box<[u8]>,
}

impl Mapper1 {
    pub fn new(rom_header: &RomHeader) -> Self {
        Mapper1 {
            prg_ram: {
                let size = rom_header.prg_ram_size as usize * 8192;
                vec![0; size].into_boxed_slice()
            },
            prg_rom: {
                let size = rom_header.prg_rom_size as usize * 16384;
                vec![0; size].into_boxed_slice()
            },
            chr: {
                let size = rom_header.chr_rom_size as usize * 8192;
                vec![0; size].into_boxed_slice()
            },
        }
    }
}

impl Mapper for Mapper1 {
    fn prg_rom_read(&self, addr: u16) -> u8 {}

    fn prg_rom_write(&mut self, addr: u16, val: u8) {}

    fn chr_rom_read(&self, addr: u16) -> u8 {}

    fn chr_rom_write(&mut self, addr: u16, val: u8) {}
}

我在其中尝试定义一个特征，在多个结构上实现该特征，然后具有一个函数，其中 returns 这些结构之一。这甚至可能吗？

我收到编译器错误：

expected `_`,
    found `mapper::Mapper1`
(expected type parameter,
    found struct `mapper::Mapper1`) [E0308]
src/mapper.rs:11     match rom_header.mapper_number {
src/mapper.rs:12         0 => Mapper1::new(rom_header),
src/mapper.rs:13         _ => panic!("Unsupported mapper: {:#}", rom_header.mapper_number),
src/mapper.rs:14     }
src/mapper.rs:11:5: 14:6 help: run `rustc --explain E0308` to see a detailed explanation
src/mapper.rs:12:14: 12:38 note: match arm with an incompatible type
src/mapper.rs:12         0 => Mapper1::new(rom_header),
                              ^~~~~~~~~~~~~~~~~~~~~~~~

Answer 1

pub fn choose_mapper<M: Mapper>(rom_header: &RomHeader) -> M 表示函数 returns M，它是一些（不是任何）Mapper。泛型由调用者提供（明确地使用 choose_mapper::<SomeM>(foo) 或大部分时间从上下文推断，如 let bar: SomeM = choose_mapper(foo)）。

您的函数尝试 return 一个 Mapper1（这是一些 M，但不一定与调用者想要的相同）。

您应该将签名更改为

pub fn choose_mapper(rom_header: &RomHeader) -> Box<Mapper>;

它允许函数选择它 returns.

有趣的是 a very active RFC 允许函数选择它 return 的内容（通常称为 impl Trait，尽管尚未选择语法）。

难以理解 Rust 中的特征和泛型

Having trouble understanding traits and generics in Rust

generics

rust

traits