Type inference in Go - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Go code using type inference?

Consider the following Go program. What will it print when run?

Go

package main
import "fmt"
func main() {
    x := 10
    y := 3.5
    z := x + int(y)
    fmt.Println(z)
}

A Error: cannot add int and float64

B 13.5

C 10

D 13

💡 Hint

Remember Go requires explicit conversion between int and float64 types.

In Go, you cannot add int and float64 directly. The float64 y is converted to int before addition, so 3.5 becomes 3, and 10 + 3 = 13.

❓ Predict Output

intermediate

What is the type of variable 'a' inferred by Go?

Look at this Go code snippet. What is the inferred type of variable 'a'?

Go

package main
func main() {
    a := 'G'
}

A rune

B string

C int

D byte

💡 Hint

Single quotes in Go denote a rune literal.

In Go, single quotes denote a rune literal, which is an alias for int32. So 'G' is of type rune.

🔧 Debug

advanced

Why does this Go code cause a compilation error?

Examine the code below. Why does it fail to compile?

Go

package main
func main() {
    var x = 5
    x = "hello"
}

A Variable x is not declared

B Missing semicolon after variable declaration

C Cannot assign string to variable x of inferred type int

D Cannot assign int to string variable

💡 Hint

Check the type inferred for x and what value is assigned later.

Variable x is inferred as int from the initial value 5. Assigning a string "hello" to it later causes a type mismatch error.

❓ Predict Output

advanced

What is the output of this Go code using type inference with constants?

What will this Go program print?

Go

package main
import "fmt"
func main() {
    const pi = 3.14
    radius := 5
    area := pi * float64(radius) * float64(radius)
    fmt.Printf("%.2f", area)
}

A 78.50

B 78

C Error: cannot convert radius to float64

D 15.7

💡 Hint

Look at how radius is converted before multiplication.

radius is int, converted to float64 for multiplication with pi (a constant float64). The area is 3.14 * 5 * 5 = 78.5, printed with two decimals.

🧠 Conceptual

expert

Which statement about Go's type inference is true?

Select the correct statement about how Go infers types for variables declared with ':='.

A Go allows the variable type to change dynamically at runtime based on assigned values.

B Go infers the variable type from the right-hand side expression at compile time and it cannot change later.

C Go infers the variable type as interface{} by default when using ':='.

D Go requires explicit type annotations even when using ':='.

💡 Hint

Think about static typing and how ':=' works in Go.

Go is statically typed. When using ':=', the compiler infers the type from the initial value and the variable's type cannot change later.