Switch expression behavior in Go - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of switch with fallthrough

What is the output of this Go program?

Go

package main
import "fmt"
func main() {
    x := 2
    switch x {
    case 1:
        fmt.Println("One")
    case 2:
        fmt.Println("Two")
        fallthrough
    case 3:
        fmt.Println("Three")
    default:
        fmt.Println("Default")
    }
}

A Two\nThree

B Two

C Two\nDefault

D Three

💡 Hint

Remember that fallthrough causes the next case to run even if it doesn't match.

The switch matches case 2, prints "Two", then fallthrough causes case 3 to run, printing "Three".

❓ Predict Output

intermediate

Switch with multiple expressions

What will this Go program print?

Go

package main
import "fmt"
func main() {
    x := 5
    switch {
    case x < 0:
        fmt.Println("Negative")
    case x == 5:
        fmt.Println("Five")
    case x > 0:
        fmt.Println("Positive")
    default:
        fmt.Println("None")
    }
}

A Five

B Positive

C Negative

D None

💡 Hint

Switch without an expression evaluates cases as booleans in order.

The switch has no expression, so it checks each case condition. x == 5 is true, so it prints "Five" and stops.

❓ Predict Output

advanced

Switch with variable initialization and matching

What is the output of this Go program?

Go

package main
import "fmt"
func main() {
    switch x := 10; {
    case x < 5:
        fmt.Println("Less than 5")
    case x > 5:
        fmt.Println("Greater than 5")
    default:
        fmt.Println("Equal to 5")
    }
}

A Less than 5

B Equal to 5

C Greater than 5

D Compilation error

💡 Hint

The variable x is declared in the switch statement and used in case conditions.

x is 10, so x > 5 is true, printing "Greater than 5".

❓ Predict Output

advanced

Switch with type assertion and interface

What will this Go program print?

Go

package main
import "fmt"
func main() {
    var i interface{} = 3.14
    switch v := i.(type) {
    case int:
        fmt.Println("int", v)
    case float64:
        fmt.Println("float64", v)
    default:
        fmt.Println("unknown")
    }
}

A Compilation error

B int 3

C unknown

D float64 3.14

💡 Hint

Type switches check the dynamic type of an interface value.

i holds a float64 value 3.14, so the case float64 matches and prints "float64 3.14".

❓ Predict Output

expert

Switch with multiple expressions and fallthrough

What is the output of this Go program?

Go

package main
import "fmt"
func main() {
    x := 1
    switch x {
    case 1, 2:
        fmt.Println("One or Two")
        fallthrough
    case 3:
        fmt.Println("Three")
    case 4:
        fmt.Println("Four")
    default:
        fmt.Println("Default")
    }
}

A One or Two

B One or Two\nThree

C Three

D One or Two\nFour

💡 Hint

fallthrough passes control to the next case regardless of matching.

Case 1 matches, prints "One or Two", fallthrough causes case 3 to run, printing "Three".

Practice

1. What does a switch statement do in Go?

easy

A. It runs all cases one after another.

B. It chooses one case to run based on a value or condition.

C. It only works with numbers, not strings.

D. It repeats the same case multiple times automatically.

Solution

1. Step 1: Understand switch purpose

   A switch statement selects one matching case based on a value or condition.

2. Step 2: Confirm behavior

   It stops after the first matching case, so only one case runs.

3. Final Answer:

   It chooses one case to run based on a value or condition. -> Option B

4. Quick Check:

   Switch picks one matching case [OK]

Hint: Switch picks only the first matching case [OK]

Common Mistakes:

• Thinking switch runs all cases
• Believing switch only works with numbers
• Assuming switch repeats cases automatically

2. Which of the following is the correct syntax for a basic switch statement in Go?

easy

A. switch (value) { case 1: doSomething() default: doDefault() }

B. switch value do { case 1: doSomething() default: doDefault() }

C. switch value then { case 1: doSomething() default: doDefault() }

D. switch value { case 1: doSomething() default: doDefault() }

Solution

1. Step 1: Recall Go switch syntax

   Go uses switch value { case ... } without parentheses or extra keywords.

2. Step 2: Check options

   switch value { case 1: doSomething() default: doDefault() } matches correct syntax; others add invalid keywords or parentheses.

3. Final Answer:

   switch value { case 1: doSomething() default: doDefault() } -> Option D

4. Quick Check:

   Go switch uses braces, no parentheses [OK]

Hint: No parentheses around switch value in Go [OK]

Common Mistakes:

• Adding parentheses around switch value
• Using 'then' or 'do' keywords incorrectly
• Misplacing braces or colons

3. What is the output of this Go code?

package main
import "fmt"
func main() {
  x := 3
  switch x {
  case 1:
    fmt.Println("One")
  case 2, 3:
    fmt.Println("Two or Three")
  default:
    fmt.Println("Other")
  }
}

medium

A. One

B. Other

C. Two or Three

D. No output

Solution

1. Step 1: Identify switch value and cases

   Variable x is 3. The switch checks cases 1, then 2 or 3, then default.

2. Step 2: Match case

   Case 2, 3 matches because x is 3, so it prints "Two or Three".

3. Final Answer:

   Two or Three -> Option C

4. Quick Check:

   3 matches case 2,3 [OK]

Hint: Check all case values including multiple ones [OK]

Common Mistakes:

• Ignoring multiple values in one case
• Choosing default instead of matching case
• Assuming case 3 is separate from case 2

4. Identify the error in this Go switch code:

package main
import "fmt"
func main() {
  x := 5
  switch {
  case x == 1:
    fmt.Println("One")
  case x == 5:
    fmt.Println("Five")
  case x == 10:
    fmt.Println("Ten")
  }
}

medium

A. No error; code runs and prints "Five"

B. Switch expression is missing a value but cases use conditions

C. Missing default case causes error

D. Case statements must be constants, not conditions

Solution

1. Step 1: Understand switch without expression

   Switch without a value allows cases to be conditions (boolean expressions).

2. Step 2: Check cases and output

   x is 5, so case x == 5 is true, printing "Five". No error occurs.

3. Final Answer:

   No error; code runs and prints "Five" -> Option A

4. Quick Check:

   Switch without value uses conditions [OK]

Hint: Switch without value uses boolean cases [OK]

Common Mistakes:

• Thinking default is mandatory
• Assuming switch must have a value
• Believing case conditions are invalid

5. Given this Go code, what will be printed?

package main
import "fmt"
func main() {
  score := 75
  switch {
  case score >= 90:
    fmt.Println("Grade A")
  case score >= 80:
    fmt.Println("Grade B")
  case score >= 70:
    fmt.Println("Grade C")
  default:
    fmt.Println("Fail")
  }
}

hard

A. Grade C

B. Grade B

C. Fail

D. Grade A

Solution

1. Step 1: Analyze switch without expression

   Switch uses conditions in cases, evaluated top to bottom.

2. Step 2: Evaluate conditions for score = 75

   score >= 90 is false, score >= 80 is false, score >= 70 is true, so prints "Grade C" and stops.

3. Final Answer:

   Grade C -> Option A

4. Quick Check:

   First true case runs, here score >= 70 [OK]

Hint: Switch stops at first true condition [OK]

Common Mistakes:

• Ignoring order of cases
• Assuming all true cases print
• Confusing >= with >