传递模块作为参数的正确方法是什么？
提问于 2018-03-14 11:54:18
我有一个模块文件(src/map/map.rs)：
type Layer = Vec<Vec<Tile>>;
pub struct Map {
height: i32,
width: i32,
data: Layer,
rooms: Vec<Rect>,
}
impl Map {
pub fn new(width: i32, height: i32) -> Self {
Map {
height: height,
width: width,
data: vec![vec![Tile::wall(); height as usize]; width as usize],
rooms: vec![Rect],
}
}
pub fn generate_with(&self, creator: module) {
creator::generate(&self)
}
}具有文件中一个函数的嵌套模块map::gen::dungeon::basic (src/map/gen/dungeon/basic.rs)：
pub fn generate(map: &mut Map) -> (Map, (i32, i32)) {}和map模块文件(src/map/mod.rs)：
mod rect;
mod tile;
mod map;
pub mod room;
pub mod gen;
pub use self::map::Map;
pub use self::rect::Rect;
pub use self::tile::Tile;像这样导入到main.rs中：
mod map;
use map::*;
use map::gen;我想像这样使用它：
let (map, (player_x, player_y)) = Map::new(MAP_WIDTH, MAP_HEIGHT).generate_with(gen::dungeon::basic);不过，我得到的错误是：
[cargo] expected value, found module 'gen::dungeon::basic': not a value [E]一个完整的有回购服务。
回答 1
Stack Overflow用户
回答已采纳
发布于 2018-03-14 13:02:34
正如注释中所述，模块不是这样的具体概念；您试图做的事情是不可能的。
相反，您可以传递可以是值的内容：
mod basic {
pub fn generate() -> u8 {
0
}
}
mod advanced {
pub fn generate() -> u8 {
42
}
}
fn play_the_game(generator: fn() -> u8) {
let dungeon = generator();
println!("{}", dungeon);
}
fn main() {
play_the_game(basic::generate);
play_the_game(advanced::generate);
}您还可以引入一个特性并将实现类型传递为泛型：
trait DungeonGenerator {
fn generate() -> u8;
}
mod basic {
use DungeonGenerator;
pub struct Basic;
impl DungeonGenerator for Basic {
fn generate() -> u8 {
0
}
}
}
mod advanced {
use DungeonGenerator;
pub struct Advanced;
impl DungeonGenerator for Advanced {
fn generate() -> u8 {
42
}
}
}
fn play_the_game<G>()
where
G: DungeonGenerator,
{
let dungeon = G::generate();
println!("{}", dungeon);
}
fn main() {
play_the_game::<basic::Basic>();
play_the_game::<advanced::Advanced>();
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/49286608
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