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Rustprogramming~5 mins

Creating threads in Rust

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Introduction

Threads let your program do many things at the same time. This helps make your program faster and more efficient.

When you want to download multiple files at once.
When you want to run a timer while doing other tasks.
When you want to handle many users in a chat app simultaneously.
When you want to perform background calculations without stopping the main program.
Syntax
Rust
use std::thread;

fn main() {
    let handle = thread::spawn(|| {
        // code to run in new thread
    });

    handle.join().unwrap();
}

thread::spawn starts a new thread and runs the code inside the closure.

handle.join() waits for the thread to finish before continuing.

Examples
This example creates a thread that prints a message, then waits for it to finish.
Rust
use std::thread;

fn main() {
    thread::spawn(|| {
        println!("Hello from a thread!");
    }).join().unwrap();
}
This thread counts from 1 to 3 and prints each number.
Rust
use std::thread;

fn main() {
    let handle = thread::spawn(|| {
        for i in 1..4 {
            println!("Counting: {}", i);
        }
    });

    handle.join().unwrap();
}
Sample Program

This program runs two loops at the same time: one in the main thread and one in a new thread. You will see their outputs mixed because they run together.

Rust
use std::thread;

fn main() {
    let handle = thread::spawn(|| {
        for i in 1..=5 {
            println!("Thread counting: {}", i);
        }
    });

    for i in 1..=3 {
        println!("Main thread counting: {}", i);
    }

    handle.join().unwrap();
}
OutputSuccess
Important Notes

Threads run independently, so their output order can change each time you run the program.

Always use join() to wait for threads to finish, or your program might end early.

Summary

Threads let your program do many things at once.

Use thread::spawn to create a new thread.

Use join() to wait for a thread to finish.

Practice

(1/5)
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
Recommended

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10 questions~5 mineasy hard