Channel creation in Go - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Go program using a channel?

Consider the following Go code that creates a channel and sends a value. What will be printed?

Go

package main
import "fmt"
func main() {
    ch := make(chan int)
    go func() {
        ch <- 42
    }()
    val := <-ch
    fmt.Println(val)
}

A 42

B 0

C Deadlock runtime panic

D Compilation error

💡 Hint

Think about how the goroutine sends the value and the main goroutine receives it.

The channel ch is created and a goroutine sends the value 42 into it. The main goroutine receives from ch and prints the value 42.

❓ Predict Output

intermediate

What happens if you create a channel without a buffer and send without a receiver?

What will happen when running this Go code?

Go

package main
func main() {
    ch := make(chan int)
    ch <- 1
}

A Deadlock runtime panic

B Compilation error

C Program prints 1

D Program exits silently

💡 Hint

Consider what happens when sending on an unbuffered channel without a receiver.

Sending on an unbuffered channel blocks until another goroutine receives. Since no receiver exists, the program deadlocks and panics.

🔧 Debug

advanced

Identify the error in this channel creation and usage

What error will this Go program produce?

Go

package main
import "fmt"
func main() {
    var ch chan int
    ch <- 5
    fmt.Println(<-ch)
}

A Program prints 5

B Compilation error: cannot send to nil channel

C Runtime panic: send on closed channel

D Deadlock runtime panic

💡 Hint

Think about the initial value of an uninitialized channel variable.

The channel ch is nil because it was declared but not initialized with make. Sending to a nil channel blocks forever, causing a deadlock panic.

❓ Predict Output

advanced

What is the output of this buffered channel example?

What will this Go program print?

Go

package main
import "fmt"
func main() {
    ch := make(chan string, 2)
    ch <- "hello"
    ch <- "world"
    fmt.Println(<-ch)
    fmt.Println(<-ch)
}

A Deadlock runtime panic

B world\nhello

C hello\nworld

D Compilation error

💡 Hint

Remember that buffered channels allow sending without immediate receiving until the buffer is full.

The buffered channel holds two strings. The sends do not block. Receiving prints the values in the order sent: "hello" then "world".

❓ Predict Output

expert

What is the length and capacity of this channel after creation?

Given this Go code, what are the length and capacity of the channel ch immediately after creation?

Go

package main
import "fmt"
func main() {
    ch := make(chan int, 5)
    ch <- 10
    ch <- 20
    fmt.Println(len(ch), cap(ch))
}

A 0 5

B 2 5

C 2 2

D Compilation error

💡 Hint

Use len to get the number of elements in the channel buffer and cap for its capacity.

The channel has capacity 5 and currently holds 2 elements, so len(ch) is 2 and cap(ch) is 5.