How do I disambiguate traits in Rust?

Question

I want to use the write_fmt method on two different types of object:

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"));
}

I get an error when using Write because they are both named the same:

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"));

Or I get an error saying the trait is not available:

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

You can call the trait method directly:

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"));
}

You can also choose to only import the trait in a smaller scope:

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"));
    }
}

---

Note that if you choose to use a smaller scope, you can also use the write! macro directly:

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");
    }
}

In either case, you should handle the Result return value.

See also:

• How to call a method when a trait and struct use the same name?

Answer 2

You can specify an alias for use:

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"));
}

If you just want to call the traits method, you can even just bring them in scope:

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

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"));
}

Be careful, this solution works when different types implement different traits with the same name. If the same type implements different traits with the same name, you must use Shepmaster's answer:

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
    }
}