Logical operators in Go - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of combined logical operators

What is the output of this Go program?

package main
import "fmt"
func main() {
  a := true
  b := false
  c := true
  fmt.Println(a && b || c)
}

Go

package main
import "fmt"
func main() {
  a := true
  b := false
  c := true
  fmt.Println(a && b || c)
}

A true

B panic at runtime

C Compilation error

D false

💡 Hint

Remember that && has higher precedence than || in Go.

The expression evaluates as (a && b) || c. Since a && b is false, and c is true, the whole expression is true.

❓ Predict Output

intermediate

Result of negation and OR operator

What will this Go code print?

package main
import "fmt"
func main() {
  x := false
  y := true
  fmt.Println(!x || y)
}

Go

package main
import "fmt"
func main() {
  x := false
  y := true
  fmt.Println(!x || y)
}

A true

B false

C Compilation error

D panic at runtime

💡 Hint

Negation (!) flips the boolean value.

Negation of false is true, so !x is true. true || y is true.

❓ Predict Output

advanced

Output of chained logical operators with variables

What is the output of this Go program?

package main
import "fmt"
func main() {
  a := false
  b := false
  c := true
  d := true
  fmt.Println(a || b && c && d)
}

Go

package main
import "fmt"
func main() {
  a := false
  b := false
  c := true
  d := true
  fmt.Println(a || b && c && d)
}

A Compilation error

B true

C false

D panic at runtime

💡 Hint

Remember operator precedence: && before ||.

b && c && d is false because b is false. a || false is false.

❓ Predict Output

advanced

Output of logical AND with short-circuit

What does this Go code print?

package main
import "fmt"
func main() {
  x := false
  y := true
  fmt.Println(x && y)
}

Go

package main
import "fmt"
func main() {
  x := false
  y := true
  fmt.Println(x && y)
}

A true

B Compilation error

C panic at runtime

D false

💡 Hint

Logical AND returns false if the first operand is false.

Since x is false, x && y is false without evaluating y.

❓ Predict Output

expert

Output of complex logical expression with parentheses

What is the output of this Go program?

package main
import "fmt"
func main() {
  a := true
  b := false
  c := true
  d := false
  fmt.Println((a || b) && (c && !d))
}

Go

package main
import "fmt"
func main() {
  a := true
  b := false
  c := true
  d := false
  fmt.Println((a || b) && (c && !d))
}

A false

B true

C Compilation error

D panic at runtime

💡 Hint

Evaluate inside parentheses first, then combine.

(a || b) is true, (c && !d) is true, so true && true is true.