Message passing concepts in Rust - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Rust message passing code?

Consider this Rust program using channels for message passing. What will it print?

Rust

use std::sync::mpsc;
use std::thread;

fn main() {
    let (tx, rx) = mpsc::channel();
    thread::spawn(move || {
        tx.send(42).unwrap();
    });
    let received = rx.recv().unwrap();
    println!("Received: {}", received);
}

A Runtime panic: channel disconnected

B Received: 0

C Compile error: cannot move `tx` into closure

D Received: 42

💡 Hint

Think about how the sender moves ownership of the transmitter into the thread and sends the value.

The sender tx is moved into the spawned thread, which sends the value 42. The receiver rx receives it and prints it.

🧠 Conceptual

intermediate

Which statement about Rust channels is true?

Choose the correct statement about Rust's std::sync::mpsc channels.

A Messages sent through channels must implement the Copy trait.

B The receiver can be cloned to allow multiple receivers.

C The sender can be cloned to allow multiple senders.

D Channels in Rust are synchronous by default.

💡 Hint

Think about which side of the channel supports cloning for multiple producers or consumers.

In Rust, the sender (tx) can be cloned to allow multiple threads to send messages. The receiver (rx) cannot be cloned.

🔧 Debug

advanced

What error does this Rust code produce?

Examine this Rust code snippet using channels. What error will it produce when compiled?

Rust

use std::sync::mpsc;

fn main() {
    let (tx, rx) = mpsc::channel();
    tx.send(10).unwrap();
    let val = rx.recv().unwrap();
    let val2 = rx.recv().unwrap();
    println!("{} {}", val, val2);
}

A Deadlock error at runtime

B Runtime panic: channel disconnected on second recv

C Output: 10 10

D Compile error: cannot borrow `rx` as mutable twice

💡 Hint

Consider what happens when you try to receive more messages than sent.

Only one message is sent. The first recv() gets 10. The second recv() blocks indefinitely because the sender tx is still in scope (the main thread is blocked on recv(), preventing tx from being dropped), resulting in a deadlock.

📝 Syntax

advanced

Which option fixes the syntax error in this Rust message passing code?

Identify the correct fix for the syntax error in this Rust code snippet:

let (tx, rx) = mpsc::channel();
thread::spawn(|| {
    tx.send(5).unwrap();
});

A Add semicolon after channel creation: <code>let (tx, rx) = mpsc::channel();;</code>

B Change closure to move: <code>thread::spawn(move || { tx.send(5).unwrap(); });</code>

C Declare tx as mutable: <code>let mut tx = mpsc::channel().0;</code>

D Use recv instead of send: <code>tx.recv(5).unwrap();</code>

💡 Hint

Think about ownership when moving variables into threads.

The closure needs to take ownership of tx to send the message inside the new thread. Adding move fixes the syntax error.

🚀 Application

expert

How many messages are received in this Rust program?

Given this Rust program using multiple senders, how many messages will the receiver get?

Rust

use std::sync::mpsc;
use std::thread;

fn main() {
    let (tx, rx) = mpsc::channel();
    for i in 0..3 {
        let tx_clone = tx.clone();
        thread::spawn(move || {
            tx_clone.send(i).unwrap();
        });
    }
    drop(tx);
    let mut count = 0;
    while let Ok(_) = rx.recv() {
        count += 1;
    }
    println!("Received {} messages", count);
}

A 3

B 4

C 0

D 1

💡 Hint

Consider how many senders send messages and when the original sender is dropped.

Three cloned senders each send one message. The original sender is dropped to close the channel. The receiver gets exactly 3 messages.

Practice

1. What is the main purpose of message passing in Rust?

easy

A. To share memory directly between threads

B. To speed up program execution by skipping safety checks

C. To safely send data between parts of a program without sharing memory

D. To write data to files asynchronously

Solution

1. Step 1: Understand message passing concept

   Message passing allows different parts of a program to communicate by sending data safely without sharing memory directly.

2. Step 2: Compare options with concept

   Only To safely send data between parts of a program without sharing memory correctly describes this safe data transfer without shared memory. Other options describe unrelated or incorrect behaviors.

3. Final Answer:

   To safely send data between parts of a program without sharing memory -> Option C

4. Quick Check:

   Message passing = safe data transfer without shared memory [OK]

Hint: Message passing means no shared memory, just safe data send [OK]

Common Mistakes:

• Thinking message passing shares memory directly
• Confusing message passing with file I/O
• Assuming message passing skips safety checks

2. Which of the following is the correct way to create a channel for message passing in Rust?

easy

A. let (tx, rx) = mpsc::channel();

B. let (tx, rx) = channel::mpsc();

C. let (tx, rx) = mpsc.channel();

D. let (tx, rx) = mpsc_channel();

Solution

1. Step 1: Recall Rust syntax for creating channels

   Rust's standard library uses mpsc::channel() function to create a sender and receiver pair.

2. Step 2: Check each option's syntax

   Only let (tx, rx) = mpsc::channel(); uses the correct syntax: mpsc::channel(). Others have wrong function names or syntax errors.

3. Final Answer:

   let (tx, rx) = mpsc::channel(); -> Option A

4. Quick Check:

   Correct channel creation syntax = let (tx, rx) = mpsc::channel(); [OK]

Hint: Use mpsc::channel() to get sender and receiver [OK]

Common Mistakes:

• Using dot instead of double colon (::)
• Swapping function and module names
• Using incorrect function names like mpsc_channel

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

use std::sync::mpsc;
use std::thread;

fn main() {
    let (tx, rx) = mpsc::channel();
    thread::spawn(move || {
        tx.send(10).unwrap();
    });
    let received = rx.recv().unwrap();
    println!("{}", received);
}

medium

A. Compilation error

B. Thread panicked

C. No output

D. 10

Solution

1. Step 1: Understand the code flow

   A channel is created, a thread sends the value 10 through the sender, and the main thread receives it and prints it.

2. Step 2: Predict the output

   The receiver gets 10 and prints it, so the output is "10".

3. Final Answer:

   10 -> Option D

4. Quick Check:

   Sent 10, received 10, printed 10 [OK]

Hint: Sender sends 10, receiver prints 10 [OK]

Common Mistakes:

• Thinking thread panics due to move keyword
• Assuming recv() blocks forever
• Expecting compilation error due to unwrap

4. Identify the error in this Rust code using message passing:

use std::sync::mpsc;
use std::thread;

fn main() {
    let (tx, rx) = mpsc::channel();
    thread::spawn(|| {
        tx.send(5).unwrap();
    });
    println!("{}", rx.recv().unwrap());
}

medium

A. tx is moved into the thread but not declared as move

B. rx cannot receive messages from another thread

C. send() returns a Result that is not handled

D. channel() returns a tuple, not two variables

Solution

1. Step 1: Analyze thread closure and tx usage

   The closure uses tx but does not declare move, so tx cannot be moved into the thread.

2. Step 2: Understand Rust's ownership rules

   Without move keyword, tx cannot be captured by the thread closure, causing a compile error.

3. Final Answer:

   tx is moved into the thread but not declared as move -> Option A

4. Quick Check:

   Missing move keyword for tx in thread closure = tx is moved into the thread but not declared as move [OK]

Hint: Use move keyword to transfer ownership into thread [OK]

Common Mistakes:

• Ignoring ownership rules for thread closures
• Thinking recv() can't get messages from other threads
• Assuming send() Result can be ignored without unwrap

5. You want to send multiple messages from one thread to another using Rust's mpsc channel. Which approach correctly allows sending multiple messages without cloning the sender each time?

hard

A. Create a new channel for each message

B. Use a single sender and send messages sequentially before dropping it

C. Clone the sender once and use the clone to send all messages

D. Use a receiver to send messages back to the sender

Solution

1. Step 1: Understand sender usage in mpsc

   A single sender can send multiple messages sequentially before it is dropped; cloning is only needed if multiple threads send concurrently.

2. Step 2: Evaluate options for sending multiple messages

   Use a single sender and send messages sequentially before dropping it correctly uses one sender to send multiple messages. Clone the sender once and use the clone to send all messages clones unnecessarily. Create a new channel for each message creates many channels, which is inefficient. Use a receiver to send messages back to the sender misuses receiver as sender.

3. Final Answer:

   Use a single sender and send messages sequentially before dropping it -> Option B

4. Quick Check:

   Single sender can send multiple messages sequentially [OK]

Hint: One sender can send many messages sequentially without cloning [OK]

Common Mistakes:

• Cloning sender unnecessarily for single-threaded sends
• Creating multiple channels instead of reusing one
• Confusing receiver with sender roles