Short variable declaration in Go - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of short variable declaration with multiple variables

What is the output of this Go program?

Go

package main
import "fmt"
func main() {
    a, b := 5, 10
    a, b = b, a
    fmt.Println(a, b)
}

A 5 10

B 0 0

C 10 5

D Compilation error

💡 Hint

Remember that := declares and initializes variables, and you can swap values using multiple assignment.

The variables a and b are declared and initialized with 5 and 10. Then their values are swapped using a, b = b, a. So the output is 10 5.

❓ Predict Output

intermediate

Short variable declaration inside if statement

What will this Go program print?

Go

package main
import "fmt"
func main() {
    if x := 7; x > 5 {
        fmt.Println(x)
    } else {
        fmt.Println(0)
    }
}

A No output

B 0

C Compilation error

D 7

💡 Hint

The variable x is declared and initialized in the if statement condition.

The short variable declaration x := 7 happens inside the if statement. Since 7 > 5 is true, it prints 7.

❓ Predict Output

advanced

Short variable declaration with redeclaration and assignment

What is the output of this Go program?

Go

package main
import "fmt"
func main() {
    x := 10
    x, y := 20, 30
    fmt.Println(x, y)
}

A 10 30

B 20 30

C Compilation error

D 10 0

💡 Hint

Short variable declaration can redeclare variables if at least one new variable is declared.

The first line declares x=10. The second line redeclares x and declares y. x is updated to 20, y is 30. So output is 20 30.

❓ Predict Output

advanced

Short variable declaration with blank identifier

What will this Go program print?

Go

package main
import "fmt"
func main() {
    x, _ := 5, 10
    fmt.Println(x)
}

A 5

B 10

C 0

D Compilation error

💡 Hint

The blank identifier _ ignores the second value.

The variable x is assigned 5, the second value 10 is discarded using _. So printing x outputs 5.

🧠 Conceptual

expert

Short variable declaration behavior in nested scopes

Consider this Go code snippet. What is the value of x printed by the program?

Go

package main
import "fmt"
func main() {
    x := 1
    {
        x := 2
        {
            x := 3
            fmt.Println(x)
        }
    }
}

A 3

B 2

C 1

D Compilation error

💡 Hint

Each nested block can declare a new variable with the same name, shadowing outer ones.

Each block declares a new x. The innermost x is 3, so fmt.Println prints 3.