抽象RUST中的可变不可变引用

因为在这两种情况下都要处理引用，所以可以定义一个泛型函数，其中T可以是&Vec<u8>或&mut Vec<u8>.所以你可以这样做:

fn get_mem<T>(address: u32, rom: T, ram: T) -> Result<T, RiscvError> {
    // ...

    match address {
        addr if (rom_start..rom_end).contains(&addr) => Ok(rom),
        addr if (ram_start..ram_end).contains(&addr) => Ok(ram),
        addr => Err(RiscvError::MemoryAlignmentError(addr)),
    }
}

impl Memory {
    pub fn get_mem_vec_ref(&self, address: u32) -> Result<&Vec<u8>, RiscvError> {
        // ...

        let mem_vec_ref = get_mem(address, &self.rom, &self.ram)?;

        return Ok(mem_vec_ref);
    }

    pub fn get_mem_vec_mut_ref(&mut self, address: u32) -> Result<&mut Vec<u8>, RiscvError> {
        // ...

        let mem_vec_ref = get_mem(address, &mut self.rom, &mut self.ram)?;

        return Ok(mem_vec_ref);
    }
}

现在，显然您需要修改get_mem()来说明rom_start、rom_end、ram_start、ram_end.如果您希望避免将get_mem()个字段传递给get_mem()个，那么引入一个新类型来处理地址可能是值得的，例如:

struct Addr {
    // ...
}

impl Addr {
    fn get_mem<T>(&self, rom: T, ram: T) -> Result<T, RiscvError> {
        // ...
    }
}