Question

我尝试编写泛型时非常难以理解，Rust的特质继续让我无法理解。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) {}
}

我尝试定义一个特征，在几个结构上创建该特征，然后有一个返回其中一个结构的函数。这甚至可能吗？

我收到编译错误：

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表示函数返回M，这是一些（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为let bar: SomeM = choose_mapper(foo)）。

您的函数尝试返回Mapper1（有些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>;

允许函数选择它返回的内容。

有趣的是，a very active RFC允许函数选择它返回的内容（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

1 个答案: