Sending and receiving values in Go - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to send the value 5 into the channel.

Go

ch := make(chan int)
go func() {
    ch[1] 5
}()

Drag options to blanks, or click blank then click option'

A<-chan<-

B<-chan

C<-

D<-chan<-chan

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<-' on the wrong side of the channel.

Confusing channel types with send/receive operators.

To send a value into a channel in Go, use the syntax 'channel <- value'.

2fill in blank

medium

Complete the code to receive a value from the channel and store it in variable x.

Go

ch := make(chan int)
go func() {
    ch <- 10
}()
x := [1] ch

Drag options to blanks, or click blank then click option'

Asend

Bmake

Cchan

D<-

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'chan' keyword instead of receive operator.

Using send operator '<-' on the wrong side.

To receive a value from a channel, use '<-channel'.

3fill in blank

hard

Fix the error in the code to correctly send the string "hello" into the channel.

Go

ch := make(chan string)
go func() {
    [1] <- "hello"
}()

Drag options to blanks, or click blank then click option'

A"ch"

Bch

Cchan

Dmake

Attempts:

3 left

💡 Hint

Common Mistakes

Using the channel name as a string.

Using 'chan' keyword instead of the channel variable.

To send a value into a channel, use the channel variable name followed by '<- value'.

4fill in blank

hard

Fill both blanks to create a channel of integers and send the value 42 into it.

Go

ch := make([1])
go func() {
    [2] <- 42
}()

Drag options to blanks, or click blank then click option'

Achan int

Bchan string

Cch

Dchan bool

Attempts:

3 left

💡 Hint

Common Mistakes

Creating a channel of wrong type.

Using wrong variable name to send the value.

Use 'make(chan int)' to create an integer channel and 'ch <- 42' to send the value.

5fill in blank

hard

Fill all three blanks to receive a value from the channel and print it.

Go

ch := make([1])
go func() {
    ch <- "GoLang"
}()
value := [2] [3]
fmt.Println(value)

Drag options to blanks, or click blank then click option'

Achan string

B<-

Cch

Dchan int

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong channel type.

Mixing up send and receive operators.

Using wrong variable name for the channel.

Create a string channel with 'chan string', receive with '<- ch', then print the value.

Practice

1. What is the correct way to send a value 5 to a channel named ch in Go?

easy

A. ch <- 5

B. 5 <- ch

C. send(ch, 5)

D. ch = 5

Solution

1. Step 1: Understand channel send syntax

   In Go, to send a value to a channel, you write channel <- value.

2. Step 2: Apply syntax to given variables

   Here, the channel is ch and the value is 5, so the correct send is ch <- 5.

3. Final Answer:

   ch <- 5 -> Option A

4. Quick Check:

   Send syntax = channel <- value [OK]

Hint: Remember: channel <- value sends data [OK]

Common Mistakes:

• Reversing send syntax as value <- channel
• Using assignment operator instead of send
• Calling a non-existent send function

2. Which of the following is the correct syntax to receive a value from a channel ch and store it in variable v?

easy

A. v := ch <-

B. v = ch <-

C. v = ch.receive()

D. v := <- ch

Solution

1. Step 1: Recall channel receive syntax

   To receive a value from a channel, Go uses value := <- channel.

2. Step 2: Match syntax with variables

   Here, v receives from ch, so v := <- ch is correct.

3. Final Answer:

   v := <- ch -> Option D

4. Quick Check:

   Receive syntax = v := <- ch [OK]

Hint: Use := <- channel to receive value [OK]

Common Mistakes:

• Placing <- after channel name
• Using method calls like receive()
• Using assignment = instead of := for declaration

3. What will be the output of the following Go program?

package main
import "fmt"
func main() {
    ch := make(chan int)
    go func() {
        ch <- 10
    }()
    val := <-ch
    fmt.Println(val)
}

medium

A. 0

B. Compilation error

C. 10

D. Deadlock error

Solution

1. Step 1: Understand channel creation and goroutine

   The channel ch is created and a goroutine sends 10 to it.

2. Step 2: Receive value and print

   The main goroutine receives the value from ch and prints it.

3. Final Answer:

   10 -> Option C

4. Quick Check:

   Send and receive work correctly, output = 10 [OK]

Hint: Send in goroutine, receive in main prints sent value [OK]

Common Mistakes:

• Expecting zero value instead of sent value
• Thinking channel blocks forever causing deadlock
• Confusing send and receive syntax

4. Identify the error in this Go code snippet:

package main
func main() {
    ch := make(chan int)
    ch <- 5
    val := <- ch
    println(val)
}

medium

A. Channel is not created properly

B. Sending to channel blocks forever causing deadlock

C. Receiving from channel is incorrect syntax

D. Variable val is not declared

Solution

1. Step 1: Analyze channel usage without goroutine

   The code sends to ch but no goroutine is receiving yet, so send blocks forever.

2. Step 2: Understand deadlock cause

   Since the send blocks and main goroutine waits, program deadlocks before receiving.

3. Final Answer:

   Sending to channel blocks forever causing deadlock -> Option B

4. Quick Check:

   Send blocks without concurrent receiver [OK]

Hint: Send blocks if no goroutine receives concurrently [OK]

Common Mistakes:

• Assuming send works without concurrent receive
• Thinking channel creation is wrong
• Confusing syntax errors with runtime deadlock

5. You want to send multiple values 1, 2, 3 to a channel ch and receive them in order. Which code snippet correctly does this without causing deadlock?

hard

A. ch := make(chan int)
go func() {
ch <- 1
ch <- 2
ch <- 3
}()
for i := 0; i < 3; i++ {
fmt.Println(<-ch)
}

B. ch := make(chan int)
ch <- 1
ch <- 2
ch <- 3
for i := 0; i < 3; i++ {
fmt.Println(<-ch)
}

C. ch := make(chan int, 1)
ch <- 1
ch <- 2
ch <- 3
for i := 0; i < 3; i++ {
fmt.Println(<-ch)
}

D. ch := make(chan int)
for i := 1; i <= 3; i++ {
ch <- i
}
for i := 0; i < 3; i++ {
fmt.Println(<-ch)
}

Solution

1. Step 1: Check for deadlock prevention

   ch := make(chan int)
   go func() {
       ch <- 1
       ch <- 2
       ch <- 3
   }()
   for i := 0; i < 3; i++ {
       fmt.Println(<-ch)
   }

   uses a goroutine to send values, so sending won't block the main goroutine.

2. Step 2: Confirm receiving order and printing

   Main goroutine receives three values in order and prints them correctly.

3. Final Answer:

   This code runs without deadlock and prints 1, 2, 3 -> Option A

4. Quick Check:

   Use goroutine to send multiple values safely [OK]

Hint: Send in goroutine or use buffered channel to avoid deadlock [OK]

Common Mistakes:

• Sending multiple values on unbuffered channel without goroutine
• Assuming unbuffered channel allows multiple sends without receive
• Forgetting to import fmt for printing