Goroutine creation - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of simple goroutine execution

What is the output of this Go program?

Go

package main
import (
	"fmt"
	"time"
)
func main() {
	go fmt.Println("Hello from goroutine")
	time.Sleep(100 * time.Millisecond)
	fmt.Println("Main function finished")
}

A

Hello from goroutine
Main function finished

B

Main function finished
Hello from goroutine

C Hello from goroutine

D Main function finished

💡 Hint

Think about how goroutines run concurrently and why we use Sleep here.

The goroutine prints first, but since goroutines run concurrently, the main function waits 100ms to let it finish before printing its own message.

❓ Predict Output

intermediate

Goroutine variable capture behavior

What will be printed by this Go program?

Go

package main
import (
	"fmt"
	"time"
)
func main() {
	for i := 0; i < 3; i++ {
		go func() {
			fmt.Println(i)
		}()
	}
	time.Sleep(100 * time.Millisecond)
}

A

3
3
3

B

0
1
2

C

0
0
0

D

2
2
2

💡 Hint

Consider how the loop variable is captured by the goroutine function.

The anonymous function captures the variable i by reference, so when the goroutines run, i has already reached 3.

🔧 Debug

advanced

Identify the error in goroutine creation

What error will this Go program produce when run?

Go

package main
func main() {
	go func() {
		println("Running")
	}()
}

A Runtime panic: goroutine not started

B Syntax error: missing import fmt

C Deadlock detected

D No output, program exits immediately

💡 Hint

Think about what happens to the main function after starting a goroutine.

The main function exits immediately after starting the goroutine, so the goroutine may not run before program exit.

❓ Predict Output

advanced

Goroutine with argument capture

What is the output of this Go program?

Go

package main
import (
	"fmt"
	"time"
)
func main() {
	for i := 0; i < 3; i++ {
		go func(n int) {
			fmt.Println(n)
		}(i)
	}
	time.Sleep(100 * time.Millisecond)
}

A

2
2
2

B

3
3
3

C

0
1
2

D

0
0
0

💡 Hint

Look at how the loop variable is passed as an argument to the goroutine function.

Passing i as an argument copies its value at each iteration, so each goroutine prints its own value.

🧠 Conceptual

expert

Goroutine scheduling and output order

Given this Go program, which output is guaranteed?

Go

package main
import (
	"fmt"
	"sync"
)
func main() {
	var wg sync.WaitGroup
	wg.Add(2)
	go func() {
		defer wg.Done()
		fmt.Println("First")
	}()
	go func() {
		defer wg.Done()
		fmt.Println("Second")
	}()
	wg.Wait()
}

A

First
Second

B Either "First\nSecond" or "Second\nFirst"

C

Second
First

D No output

💡 Hint

Consider how goroutines are scheduled and the use of WaitGroup.

The WaitGroup ensures both goroutines finish before main exits, but their print order is not guaranteed.