Creating threads in Rust - Performance & Efficiency

Time Complexity: Creating threads

O(n)

Understanding Time Complexity

When we create threads in Rust, we want to know how the time to start them changes as we create more threads.

We ask: How does the work grow when we add more threads?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

use std::thread;

fn main() {
    let n = 100usize;
    let mut handles = vec![];
    for i in 0..n {
        let handle = thread::spawn(move || {
            println!("Thread {} is running", i);
        });
        handles.push(handle);
    }
    for handle in handles {
        handle.join().unwrap();
    }
}

This code creates n threads, each printing a message, then waits for all to finish.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Loop creating and starting threads n times.
• How many times: Exactly once per thread, so n times.

How Execution Grows With Input

Each new thread adds a fixed amount of work to start and run.

Input Size (n)	Approx. Operations
10	About 10 thread starts and joins
100	About 100 thread starts and joins
1000	About 1000 thread starts and joins

Pattern observation: The work grows directly with the number of threads.

Final Time Complexity

Time Complexity: O(n)

This means the time to create and wait for threads grows in a straight line as we add more threads.

Common Mistake

[X] Wrong: "Creating threads is instant and does not add time as we add more."

[OK] Correct: Each thread takes some time to start and finish, so more threads mean more total time.

Interview Connect

Understanding how thread creation time grows helps you write programs that use threads wisely and explain your choices clearly.

Self-Check

"What if we reused a fixed number of threads instead of creating new ones each time? How would the time complexity change?"

Practice

1. What does thread::spawn do in Rust?

easy

A. It stops all threads immediately.

B. It pauses the current thread for a set time.

C. It creates a new thread to run code concurrently.

D. It merges two threads into one.

Solution

1. Step 1: Understand the purpose of thread::spawn

   This function is used to start a new thread that runs code independently from the main thread.

2. Step 2: Compare options with the function's role

   Only "It creates a new thread to run code concurrently." correctly describes creating a new thread. Other options describe unrelated actions.

3. Final Answer:

   It creates a new thread to run code concurrently. -> Option C

4. Quick Check:

   thread::spawn creates threads [OK]

Hint: Remember: spawn means start a new thread [OK]

Common Mistakes:

• Confusing spawn with pausing or stopping threads
• Thinking spawn merges threads
• Assuming spawn waits for thread completion

2. Which of the following is the correct syntax to create a new thread in Rust?

easy

A. let handle = thread::spawn(|| { println!("Hello"); });

B. let handle = spawn::thread(|| { println!("Hello"); });

C. thread.spawn(|| println!("Hello"));

D. thread::create(|| { println!("Hello"); });

Solution

1. Step 1: Recall the correct module and function

   The correct function to create a thread is thread::spawn from the std::thread module.

2. Step 2: Check syntax correctness

   let handle = thread::spawn(|| { println!("Hello"); }); uses the correct syntax: thread::spawn(|| { ... }). Other options use wrong module names or functions.

3. Final Answer:

   let handle = thread::spawn(|| { println!("Hello"); }); -> Option A

4. Quick Check:

   Correct syntax uses thread::spawn [OK]

Hint: Use thread::spawn with a closure to start threads [OK]

Common Mistakes:

• Swapping module and function names
• Missing double colons (::) in path
• Using non-existent functions like create()

3. What will be the output of this Rust code?

use std::thread;

fn main() {
    let handle = thread::spawn(|| {
        println!("Inside thread");
    });
    println!("Outside thread");
    handle.join().unwrap();
}

medium

A. Inside thread
Outside thread

B. Outside thread
Inside thread

C. Only Inside thread

D. Only Outside thread

Solution

1. Step 1: Understand thread execution order

   The main thread prints "Outside thread" immediately. The spawned thread runs concurrently and prints "Inside thread".

2. Step 2: Consider join() effect

   The join() waits for the spawned thread to finish, but does not change the order of prints already done.

3. Final Answer:

   Outside thread Inside thread -> Option B

4. Quick Check:

   Main prints first, spawned thread prints after [OK]

Hint: Main thread prints first, spawned thread runs concurrently [OK]

Common Mistakes:

• Assuming spawned thread always prints first
• Ignoring that join() waits but doesn't reorder prints
• Thinking only one print appears

4. Identify the error in this Rust code that creates a thread:

use std::thread;

fn main() {
    let handle = thread::spawn(|| {
        println!("Hello from thread");
    });
    handle.join();
}

medium

A. println! macro cannot be used inside threads.

B. thread::spawn requires a named function, not a closure.

C. join() should be called before thread::spawn.

D. Missing call to unwrap() on join() result.

Solution

1. Step 1: Check join() usage

   The join() method returns a Result that should be handled, usually by calling unwrap() to catch panics.

2. Step 2: Verify other statements

   The other options are incorrect: println! macro works inside threads, thread::spawn accepts closures, and join() is called after spawn.

3. Final Answer:

   Missing call to unwrap() on join() result. -> Option D

4. Quick Check:

   Always unwrap join() result to handle errors [OK]

Hint: Always unwrap join() to handle thread errors [OK]

Common Mistakes:

• Ignoring join()'s Result return
• Thinking closures can't be used in spawn
• Calling join() before spawn

5. You want to create 3 threads that each print their thread number (1, 2, 3). Which code correctly does this without data races?

use std::thread;

fn main() {
    for i in 1..=3 {
        thread::spawn(|| {
            println!("Thread number: {}", i);
        });
    }
}

hard

A. Move i into closure: thread::spawn(move || { println!("Thread number: {}", i); });

B. Use thread::spawn(|| { println!("Thread number: {}", i); }); as is.

C. Declare i as static before loop.

D. Use thread::spawn(|| println!("Thread number: {}", i)); without move.

Solution

1. Step 1: Understand closure capturing in threads

   The variable i changes each loop iteration. Without move, the closure captures i by reference, causing data races.

2. Step 2: Use move to capture i by value

   Adding move moves the current value of i into the closure, making each thread print its own number safely.

3. Final Answer:

   Move i into closure: thread::spawn(move || { println!("Thread number: {}", i); }); -> Option A

4. Quick Check:

   Use move closures to capture loop variables safely [OK]

Hint: Use move closures to capture loop variables in threads [OK]

Common Mistakes:

• Not using move causes all threads to print last i value
• Trying to declare i as static (not allowed here)
• Assuming closure captures by value without move