The Question Mark `?` Operator in Rust

Alex Garella

16th November 2023

Rust, renowned for its safety and performance, introduces several features to ensure robust error handling. Among these, the Question Mark ? operator is a crucial tool for Rustaceans. It simplifies error handling, making code more readable and concise.

What is the ? Operator?

In Rust, error handling is often done using the Result type, which can be either Ok(T) representing success or Err(E) representing failure.

The ? operator is a shorthand for handling these Result types. When you append ? to a Result value, it automatically handles the error. If the Result is Ok, the value inside Ok is returned, and if it's Err, the error is returned from the whole function.

Basic Usage of ?

Let's look at a basic example of reading a file and transforming its contents to uppercase in Rust using the ? operator:

use std::fs::read_to_string; use std::io::Result; fn read_and_transform_to_uppercase(file_path: &str) -> Result<String> { let contents: String = read_to_string(file_path)?; Ok(contents.to_uppercase()) } fn main() { match read_and_transform_to_uppercase("example.txt") { Ok(contents) => println!("File contents:\n{}", contents), Err(e) => println!("Error reading file: {}", e), } }

In our read_and_transform_to_uppercase function, read_to_string followed by ? attempts to open a file. If successful, it proceeds, but if it fails (e.g., file not found), it returns the error immediately. We can then handle the success and failure cases explicitly with pattern matching in the main function.

Note that the use of ? allows us to separate the business logic from the error handing logic. Instead of handling the error and then applying the transformation, we can directly access the value, apply the transformation and move the error handling logic outside the function. Resulting in more concise and easier to read code.

Chaining Calls with ?

The ? operator is particularly useful for chaining multiple calls that return Result.

Consider this example where we chain multiple operations:

use std::fs::read_to_string; use std::fs::File; use std::io::prelude::*; use std::io::Result; fn read_transform_write(file_path_in: &str, file_path_out: &str) -> Result<()> { let contents = read_to_string(file_path_in)?; let transformed_content = contents.to_uppercase(); let mut file = File::create(file_path_out)?; file.write_all(transformed_content.as_bytes()) } fn main() { match read_transform_write("example.txt", "output.txt") { Ok(_) => println!("Success reading, transforming and writing file"), Err(e) => println!("Error reading file: {}", e), } }

In this example read_transform_write reads from the input path, transforms the contents to uppercase and writes to the file at the output path.

Here, read_to_string? and File.create? are chained. If any of these operations fail, the error is returned immediately.

Note that there is no need for the ? operator after the file.write_all expression. This is because it is the last operation in the function and we want to return its result directly.

As we can see from the example, the ? operator allows us to combine multiple operations that can produce an error within a function without having to handle each error explicitly. Instead we can handle all the errors in one go in the main function.

Using ? in Different Contexts

The ? operator can be used in functions that return Result. However, it cannot be used in main directly unless main is defined to return a Result.

Here's how you can use ? in main:

use std::fs::read_to_string; use std::error::Error; fn main() -> Result<(), Box<dyn Error>> { let content = read_to_string("example.txt")?; println!("File content: {}", content); Ok(()) }

This change allows the use of ? in main by defining the return type as Result<(), Box<dyn Error>>.

Error Conversion with ?

The ? operator automatically converts the error to the function's return type error using the From trait. This is handy when dealing with functions that return different error types.

use std::error::Error; use std::fs::read_to_string; fn read_integer_from_file(file_path: &str) -> Result<i32, Box<dyn Error>> { let contents = read_to_string(file_path)?; let num: i32 = contents.trim().parse()?; Ok(num) } fn main() { match read_integer_from_file("example.txt") { Ok(res) => println!("Successfully read number: {res}"), Err(e) => println!("Error reading file: {e}"), } }

Here, contents.trim().parse()? may produce a ParseIntError, whereas read_to_string()? may produce an Error. By using ?, ParseIntError is automatically converted into a Box<dyn Error>.


The ? operator in Rust is a powerful feature that simplifies error handling, making code cleaner and more maintainable. It enables quick propagation of errors and integrates seamlessly with Rust's robust error handling model. By understanding and utilizing the ? operator, developers can write more efficient and readable Rust code.

Subscribe to receive the latest Rust jobs in your inbox

Receive a weekly overview of Rust jobs by subscribing to our mailing list

© 2024, All rights reserved.