Propagating errors with ? in Rust - Time & Space Complexity

Time Complexity: Propagating errors with ?

O(n)

Understanding Time Complexity

When using the ? operator in Rust, we want to know how it affects the time it takes for a function to run.

Specifically, we ask: does using ? add extra work as the input grows?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn read_and_parse(input: &str) -> Result {
    let num_str = input.trim();
    let num = num_str.parse::()?;
    Ok(num)
}
    

This function trims a string, tries to parse it as a number, and uses ? to return errors early.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The parse method internally processes the string characters once.
• How many times: It runs once per function call, no loops or recursion here.

How Execution Grows With Input

The time to parse grows roughly with the length of the input string, since each character is checked.

Input Size (n)	Approx. Operations
10	About 10 character checks
100	About 100 character checks
1000	About 1000 character checks

Pattern observation: The work grows linearly as the input string gets longer.

Final Time Complexity

Time Complexity: O(n)

This means the time to run grows in a straight line with the size of the input string.

Common Mistake

[X] Wrong: "Using ? adds extra loops or slows down the function significantly."

[OK] Correct: The ? operator just returns errors early without extra looping; it does not add repeated work.

Interview Connect

Understanding how error handling affects performance shows you can write clean code without guessing about speed.

Self-Check

"What if the function called parse multiple times on different parts of the input? How would the time complexity change?"

Practice

1. What does the ? operator do in Rust when used with a Result type?

easy

A. It ignores any error and always returns the success value.

B. It converts the Result into an Option.

C. It returns the error immediately if there is one, otherwise unwraps the value.

D. It retries the operation until it succeeds.

Solution

1. Step 1: Understand the ? operator behavior

   The ? operator checks if the Result is an Err. If yes, it returns the error from the current function immediately.

2. Step 2: What happens if Result is Ok?

   If the Result is Ok, it unwraps the value inside and continues execution.

3. Final Answer:

   It returns the error immediately if there is one, otherwise unwraps the value. -> Option C

4. Quick Check:

   ? returns error or unwraps value [OK]

Hint: Remember: ? returns error early or unwraps success [OK]

Common Mistakes:

• Thinking ? converts Result to Option
• Assuming ? ignores errors
• Believing ? retries operations

2. Which of the following function signatures correctly allows using the ? operator inside the function to propagate std::io::Error?

easy

A. fn read_file() -> String

B. fn read_file() -> Result

C. fn read_file() -> ()

D. fn read_file() -> Option<String>

Solution

1. Step 1: Identify return types compatible with ?

   The ? operator works only in functions returning Result or Option.

2. Step 2: Check each signature

   fn read_file() -> String returns String directly, no error type, so ? can't be used. fn read_file() -> Result returns Result, which supports ?. fn read_file() -> () returns unit (), no error handling. fn read_file() -> Option<String> returns Option, which supports ? for absence but cannot propagate std::io::Error.

3. Final Answer:

   fn read_file() -> Result -> Option B

4. Quick Check:

   Function must return Result or Option for ? [OK]

Hint: Use Result or Option return type to use ? [OK]

Common Mistakes:

• Using ? in functions returning plain types
• Forgetting to specify error type in Result
• Confusing Option and Result return types

3. What is the output of the following Rust code?

fn parse_number(s: &str) -> Result {
    let num = s.parse::<i32>()?;
    Ok(num * 2)
}

fn main() {
    match parse_number("10") {
        Ok(n) => println!("{}", n),
        Err(e) => println!("Error: {}", e),
    }
}

medium

A. 20

B. Error: invalid digit found in string

C. 10

D. Compilation error

Solution

1. Step 1: Understand parse_number function

   The function tries to parse the string "10" to an integer using s.parse::<i32>()?. Since "10" is valid, it unwraps to 10.

2. Step 2: Calculate the returned value

   It multiplies the parsed number by 2, so 10 * 2 = 20, then returns Ok(20).

3. Step 3: Check main output

   The match prints the number 20 because parsing succeeded.

4. Final Answer:

   20 -> Option A

5. Quick Check:

   Valid parse returns doubled number [OK]

Hint: Valid parse with ? returns doubled value [OK]

Common Mistakes:

• Assuming parse fails and prints error
• Confusing output with input string
• Thinking code won't compile due to ?

4. Identify the error in this Rust function using ? and fix it:

fn get_first_char(s: &str) -> Result {
    let first = s.chars().next()?;
    Ok(first)
}

medium

A. Change next() to next().ok_or("Empty string".to_string())?

B. Change return type to Option<char>

C. Replace ? with .unwrap()

D. No error, code is correct

Solution

1. Step 1: Analyze next() return type

   next() returns an Option<char>, but ? requires Result or Option compatible with function return.

2. Step 2: Fix error propagation

   Since function returns Result, convert Option to Result using next().ok_or("Empty string".to_string())? to propagate error properly.

3. Final Answer:

   Change next() to next().ok_or("Empty string".to_string())? -> Option A

4. Quick Check:

   Convert Option to Result for ? in Result function [OK]

Hint: Convert Option to Result with ok_or before ? [OK]

Common Mistakes:

• Using ? directly on Option in Result function
• Ignoring error conversion from Option to Result
• Replacing ? with .unwrap() causing panic

5. You want to write a function that reads a file and returns the number of lines. It should propagate any IO errors using ?. Which of the following implementations correctly uses ? to propagate errors?

hard

A. fn count_lines(path: &str) -> Option<usize> {
let content = std::fs::read_to_string(path)?;
Some(content.lines().count())
}

B. fn count_lines(path: &str) -> usize {
let content = std::fs::read_to_string(path)?;
content.lines().count()
}

C. fn count_lines(path: &str) -> Result {
let content = std::fs::read_to_string(path).unwrap();
Ok(content.lines().count())
}

D. fn count_lines(path: &str) -> Result {
let content = std::fs::read_to_string(path)?;
Ok(content.lines().count())
}

Solution

1. Step 1: Check function return type compatibility

   The function returns Result, so it can use ? to propagate IO errors.

2. Step 2: Analyze each option

   fn count_lines(path: &str) -> Result { let content = std::fs::read_to_string(path)?; Ok(content.lines().count()) } uses ? correctly and returns Ok with line count. fn count_lines(path: &str) -> usize { let content = std::fs::read_to_string(path)?; content.lines().count() } returns usize directly but uses ? which is invalid here. fn count_lines(path: &str) -> Result { let content = std::fs::read_to_string(path).unwrap(); Ok(content.lines().count()) } uses unwrap() which panics on error, not propagates. fn count_lines(path: &str) -> Option<usize> { let content = std::fs::read_to_string(path)?; Some(content.lines().count()) } returns Option but read_to_string returns Result, so ? cannot be used directly.

3. Final Answer:

   fn count_lines(path: &str) -> Result<usize, std::io::Error> { let content = std::fs::read_to_string(path)?; Ok(content.lines().count()) } -> Option D

4. Quick Check:

   Return Result and use ? to propagate errors [OK]

Hint: Match function return type with ? usage for error propagation [OK]

Common Mistakes:

• Using ? in functions without Result or Option return
• Using unwrap() instead of propagating errors
• Mismatching return types and error handling