如何消除Rust中的特征歧义？

Question

我想write_fmt在两种不同类型的对象上使用该方法:

use std::fmt::Write;
use std::io::Write;

fn main() {
    let mut a = String::new();
    let mut b = std::fs::File::create("test").unwrap();

    a.write_fmt(format_args!("hello"));
    b.write_fmt(format_args!("hello"));
}

我在使用时遇到错误,Write因为它们的名称相同:

error[E0252]: a trait named `Write` has already been imported in this module
 --> src/main.rs:8:5
  |
7 | use std::fmt::Write;
  |     --------------- previous import of `Write` here
8 | use std::io::Write;
  |     ^^^^^^^^^^^^^^ `Write` already imported
      a.write_fmt(format_args!("hello"));
      b.write_fmt(format_args!("hello"));

或者我得到一个错误,说该特征不可用:

error[E0599]: no method named `write_fmt` found for type `std::fs::File` in the current scope
  --> src/main.rs:76:4
   |
76 |    b.write_fmt(format_args!("hello"));
   |      ^^^^^^^^^
   |
   = help: items from traits can only be used if the trait is in scope; the following trait is implemented but not in scope, perhaps add a `use` for it:
   = help: candidate #1: `use std::io::Write;`

Answer 1

您可以直接调用trait方法:

fn main() {
    let mut a = String::new();
    let mut b = std::fs::File::create("test").unwrap();

    std::fmt::Write::write_fmt(&mut a, format_args!("hello"));
    std::io::Write::write_fmt(&mut b, format_args!("hello"));
}

您也可以选择仅在较小的范围内导入特征:

fn main() {
    let mut a = String::new();
    let mut b = std::fs::File::create("test").unwrap();

    {
        use std::fmt::Write;
        a.write_fmt(format_args!("hello"));
    }

    {
        use std::io::Write;
        b.write_fmt(format_args!("hello"));
    }
}

请注意,如果您选择使用较小的范围,也可以write!直接使用宏:

fn main() {
    let mut a = String::new();
    let mut b = std::fs::File::create("test").unwrap();

    {
        use std::fmt::Write;
        write!(a, "hello");
    }

    {
        use std::io::Write;
        write!(b, "hello");
    }
}

在任何一种情况下,您都应该处理Result返回值.

也可以看看:

• 当trait和struct使用相同的名称时如何调用方法？

Answer 2

您可以为use以下内容指定别名:

use std::fmt::Write as FmtWrite;
use std::io::Write;

fn main() {
    let mut a = String::new();
    let mut b = std::fs::File::create("test").unwrap();

    a.write_fmt(format_args!("hello"));
    b.write_fmt(format_args!("hello"));
}

请注意,当不同类型实现具有相同名称的不同特征时,此解决方案可以正常工作.如果相同类型实现具有相同名称的不同特征,则必须使用Shepmaster的答案:

mod foo {
    pub trait Trait {
        fn do_something(&self) {}
    }
}

mod bar {
    pub trait Trait {
        fn do_something(&self) {}
    }
}

pub struct Concrete {}
impl foo::Trait for Concrete {}
impl bar::Trait for Concrete {}

fn main() {
    let x = Concrete {};

    {
        use foo::Trait; // use limited to scope

        x.do_something(); // call foo::Trait::do_something
    }
    {    
        foo::Trait::do_something(&x); // complete path to disambiguate
        bar::Trait::do_something(&x); // complete path to disambiguate
    }
    {
        use foo::Trait as FooTrait;
        use bar::Trait;

        x.do_something(&x); // ERROR: multiple applicable items in scope
    }
}