如何从同级模块导入？
如何从同级模块导入？
提问于 2015-06-06 07:27:16
在src/lib.rs中，我有以下内容
extern crate opal_core;
mod functions;
mod context;
mod shader;然后在src/context.rs中，我有类似这样的东西，它试图从src/shader.rs导入符号
use opal_core::shader::Stage;
use opal_core::shader::Shader as ShaderTrait;
use opal_core::GraphicsContext as GraphicsContextTrait;
use functions::*; // this import works fine
use shader::*; // this one doesn't
pub struct GraphicsContext {
functions: Gl
}
fn shader_stage_to_int(stage: &Stage) -> u32 {
match stage {
&Stage::Vertex => VERTEX_SHADER,
&Stage::Geometry => GEOMETRY_SHADER,
&Stage::Fragment => FRAGMENT_SHADER,
}
}
impl GraphicsContextTrait for GraphicsContext {
/// Creates a shader object
fn create_shader(&self, stage: Stage, source: &str) -> Box<ShaderTrait> {
let id;
unsafe {
id = self.functions.CreateShader(shader_stage_to_int(&stage));
}
let shader = Shader {
id: id,
stage: stage,
context: self
};
Box::new(shader)
}
}问题是语句use shader::*;给出了错误unresolved import。
我正在阅读文档，他们说use语句总是从当前机箱(opal_driver_gl)的根开始，所以我认为shader::*应该导入opal_driver_gl::shader::*，但它似乎没有这样做。这里需要使用self或super关键字吗？
如果你能帮上忙，谢谢。
回答 1
Stack Overflow用户
发布于 2015-06-07 05:37:52
请注意，use的行为已从Rust 2015更改为Rust 2018。详情请参见What are the valid path roots in the use keyword?。
Rust 2018
要导入同一级别的模块，请执行以下操作：
random_file_0.rs
// Note how this is a public function. It has to be in order to be
// usable from other files (in this case `random_file_1.rs`)
pub fn do_something() -> bool {
true
}random_file_1.rs
use super::random_file_0;
#[test]
fn do_something_else() {
assert!(random_file_0::do_something());
}或另一个random_file_1.rs
use crate::random_file_0;
#[test]
fn do_something_else() {
assert!(random_file_0::do_something());
}lib.rs
mod random_file_0;
mod random_file_1;有关更多信息和示例，请参阅Rust By Example。如果这不起作用，下面是它显示的代码：
fn function() {
println!("called `function()`");
}
mod cool {
pub fn function() {
println!("called `cool::function()`");
}
}
mod my {
fn function() {
println!("called `my::function()`");
}
mod cool {
pub fn function() {
println!("called `my::cool::function()`");
}
}
pub fn indirect_call() {
// Let's access all the functions named `function` from this scope!
print!("called `my::indirect_call()`, that\n> ");
// The `self` keyword refers to the current module scope - in this case `my`.
// Calling `self::function()` and calling `function()` directly both give
// the same result, because they refer to the same function.
self::function();
function();
// We can also use `self` to access another module inside `my`:
self::cool::function();
// The `super` keyword refers to the parent scope (outside the `my` module).
super::function();
// This will bind to the `cool::function` in the *crate* scope.
// In this case the crate scope is the outermost scope.
{
use cool::function as root_function;
root_function();
}
}
}
fn main() {
my::indirect_call();
}生锈2015年
要导入同一级别的模块，请执行以下操作：
random_file_0.rs
// Note how this is a public function. It has to be in order to be
// usable from other files (in this case `random_file_1.rs`)
pub fn do_something() -> bool {
true
}random_file_1.rs
use super::random_file_0;
#[test]
fn do_something_else() {
assert!(random_file_0::do_something());
}或另一个random_file_1.rs
use ::random_file_0;
#[test]
fn do_something_else() {
assert!(random_file_0::do_something());
}lib.rs
mod random_file_0;
mod random_file_1;以下是Rust By example以前版本中的另一个示例：
fn function() {
println!("called `function()`");
}
mod my {
pub fn indirect_call() {
// Let's access all the functions named `function` from this scope
print!("called `my::indirect_call()`, that\n> ");
// `my::function` can be called directly
function();
{
// This will bind to the `cool::function` in the *crate* scope
// In this case the crate scope is the outermost scope
use cool::function as root_cool_function;
print!("> ");
root_cool_function();
}
{
// `self` refers to the current module scope, in this case: `my`
use self::cool::function as my_cool_function;
print!("> ");
my_cool_function();
}
{
// `super` refers to the parent scope, i.e. outside of the `my`
// module
use super::function as root_function;
print!("> ");
root_function();
}
}
fn function() {
println!("called `my::function()`");
}
mod cool {
pub fn function() {
println!("called `my::cool::function()`");
}
}
}
mod cool {
pub fn function() {
println!("called `cool::function()`");
}
}
fn main() {
my::indirect_call();
}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/30677258
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