Type inference in Go - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a variable with type inference.

Go

var x = [1]

Drag options to blanks, or click blank then click option'

Avar

Bint

C"Hello, Go!"

Dstring

Attempts:

3 left

💡 Hint

Common Mistakes

Using a type name instead of a value after the equals sign.

Trying to declare type explicitly when using type inference.

In Go, you can declare a variable with type inference by assigning a value directly. Here, var x = "Hello, Go!" lets Go infer that x is a string.

2fill in blank

medium

Complete the code to declare and initialize a variable using short variable declaration with type inference.

Go

count [1] 10

Drag options to blanks, or click blank then click option'

A==

B=

Cvar

D:=

Attempts:

3 left

💡 Hint

Common Mistakes

Using single equals sign which is assignment, not declaration.

Using double equals which is a comparison operator.

The short variable declaration count := 10 declares count and lets Go infer its type as int.

3fill in blank

hard

Fix the error in the code by completing the variable declaration with type inference.

Go

var pi [1] 3.14

Drag options to blanks, or click blank then click option'

A=

B:=

Cfloat64

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using := with var keyword causes syntax error.

Using == which is a comparison operator.

When using var with explicit variable name, you must use = to assign a value. Type inference happens automatically.

4fill in blank

hard

Fill both blanks to declare a variable with type inference and assign a boolean value.

Go

isReady [1] [2]

Drag options to blanks, or click blank then click option'

A:=

Btrue

Cfalse

D=

Attempts:

3 left

💡 Hint

Common Mistakes

Using = instead of := for declaration.

Using false instead of true if the goal is to assign true.

Using isReady := true declares isReady with type inferred as bool and assigns it the value true.

5fill in blank

hard

Fill both blanks to create a map with type inference and initialize it with values.

Go

scores := map[string]int{"Alice": {BLANK_2}}, "Bob": {{BLANK_2}}

Drag options to blanks, or click blank then click option'

A{

B90

C85

D}

Attempts:

3 left

💡 Hint

Common Mistakes

Omitting the opening or closing curly braces.

Using string values instead of integers for scores.

The code scores := map[string]int{"Alice": 90, "Bob": 85} declares a map with string keys and int values, using type inference for scores.

Practice

1. What does type inference mean in Go when you write x := 10 inside a function?

easy

A. You must tell Go the type of x explicitly.

B. Go automatically figures out that x is an int.

C. Go treats x as a string by default.

D. The variable x has no type.

Solution

1. Step 1: Understand the := operator in Go

   The := operator declares a variable and assigns a value, letting Go guess the type.

2. Step 2: Check the assigned value type

   Since 10 is an integer, Go infers the type of x as int automatically.

3. Final Answer:

   Go automatically figures out that x is an int. -> Option B

4. Quick Check:

   Type inference means Go guesses type from value [OK]

Hint: := means declare and infer type from value [OK]

Common Mistakes:

• Thinking you must always write the type explicitly
• Assuming variables have no type
• Believing Go defaults to string type

2. Which of the following is the correct way to declare and assign a variable with type inference inside a function in Go?

easy

A. x := 5

B. x = 5

C. var x int = 5

D. int x = 5

Solution

1. Step 1: Recall Go syntax for type inference

   Inside functions, := declares and assigns a variable with inferred type.

2. Step 2: Check each option

   var x int = 5 declares with explicit type, not inference. x = 5 misses declaration. int x = 5 is invalid Go syntax.

3. Final Answer:

   x := 5 -> Option A

4. Quick Check:

   Use := to declare and infer type [OK]

Hint: Use := inside functions for type inference [OK]

Common Mistakes:

• Using C or D which are invalid Go syntax
• Forgetting to declare variable before assignment
• Confusing var declaration with :=

3. What is the output of this Go program?

package main
import "fmt"
func main() {
    x := 3.14
    fmt.Printf("%T", x)
}

medium

A. float64

B. float32

C. int

D. string

Solution

1. Step 1: Identify the type inferred for 3.14

   In Go, untyped floating-point constants default to float64 when assigned with :=.

2. Step 2: Understand fmt.Printf with %T

   The %T verb prints the type of the variable, so it will print "float64".

3. Final Answer:

   float64 -> Option A

4. Quick Check:

   Float literals default to float64 [OK]

Hint: Float literals default to float64 type [OK]

Common Mistakes:

• Assuming float32 is default
• Thinking 3.14 is int
• Confusing %T output with value

4. What is wrong with this Go code snippet?

func main() {
    var x := 10
    fmt.Println(x)
}

medium

A. The value 10 is not a valid integer.

B. x is not declared before use.

C. You cannot use := with var keyword.

D. fmt.Println is not imported.

Solution

1. Step 1: Understand variable declaration syntax

   In Go, := is shorthand for declaration and assignment without var keyword.

2. Step 2: Identify the syntax error

   Using var and := together is invalid syntax; you must use either var x = 10 or x := 10.

3. Final Answer:

   You cannot use := with var keyword. -> Option C

4. Quick Check:

   var and := cannot be combined [OK]

Hint: Don't combine var and := in one statement [OK]

Common Mistakes:

• Mixing var and := together
• Ignoring import errors (not shown here)
• Thinking := works outside functions

5. Given this Go code, what is the type of result after execution?

func main() {
    a := 5
    b := 2.0
    result := a * int(b)
    fmt.Printf("%T", result)
}

hard

A. float64

B. string

C. float32

D. int

Solution

1. Step 1: Analyze variable types and conversion

   a is int, b is float64 (default for 2.0). The expression converts b to int before multiplication.

2. Step 2: Determine result type

   Multiplying int (a) by int (converted b) results in int type for result.

3. Final Answer:

   int -> Option D

4. Quick Check:

   int * int = int type [OK]

Hint: Type conversion affects result type [OK]

Common Mistakes:

• Assuming result is float64 without conversion
• Forgetting to convert float to int
• Confusing multiplication type rules