Defining structs in Go - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of struct field access

What is the output of this Go program?

Go

package main
import "fmt"
type Person struct {
  Name string
  Age  int
}
func main() {
  p := Person{Name: "Alice", Age: 30}
  fmt.Println(p.Name)
}

A Alice

B 30

C Person{Name: "Alice", Age: 30}

D Compilation error

💡 Hint

Look at which field is printed with fmt.Println.

The program creates a Person struct with Name "Alice" and Age 30. It prints p.Name, so output is "Alice".

❓ Predict Output

intermediate

Output when struct fields are uninitialized

What is the output of this Go program?

Go

package main
import "fmt"
type Point struct {
  X int
  Y int
}
func main() {
  var p Point
  fmt.Println(p.X, p.Y)
}

A nil nil

B Compilation error: missing initialization

C 0 0

D Random values

💡 Hint

Uninitialized int fields default to zero.

In Go, int fields default to 0 if not initialized. So p.X and p.Y print as 0 0.

🔧 Debug

advanced

Identify the error in struct initialization

What error does this Go code produce?

Go

package main
import "fmt"
type Car struct {
  Brand string
  Year  int
}
func main() {
  c := Car{Brand: "Toyota", Year: 2020}
  fmt.Println(c)
}

A Syntax error: mixed field names and values

B Compilation error: missing field Year

C Runtime panic: invalid type conversion

D No error, prints {Toyota 2020}

💡 Hint

Check how struct fields are initialized with names and values.

The struct is correctly initialized with named fields. This code compiles and prints {Toyota 2020}.

❓ Predict Output

advanced

Output of nested structs

What is the output of this Go program?

Go

package main
import "fmt"
type Address struct {
  City  string
  State string
}
type User struct {
  Name    string
  Address Address
}
func main() {
  u := User{Name: "Bob", Address: Address{City: "NYC", State: "NY"}}
  fmt.Println(u.Address.City)
}

A NY

B NYC

C {City: NYC, State: NY}

D Compilation error

💡 Hint

Look at which nested field is printed.

The program prints u.Address.City which is "NYC".

🧠 Conceptual

expert

Number of fields in an anonymous struct

How many fields does this struct have?

Go

var s = struct {
  A int
  B string
  C struct {
    D float64
    E bool
  }
}{}

A 3

B 5

C 2

D 4

💡 Hint

Count only the top-level fields of the struct.

The struct has three top-level fields: A, B, and C. The nested struct inside C is one field itself.

Practice

1. What is the purpose of defining a struct in Go?

easy

A. To create a new function

B. To group related data fields into a single unit

C. To declare a package

D. To import external libraries

Solution

1. Step 1: Understand the role of structs

   Structs are used to group related data fields together in one unit, like a blueprint for an object.

2. Step 2: Compare with other options

   Functions, packages, and imports serve different purposes unrelated to grouping data.

3. Final Answer:

   To group related data fields into a single unit -> Option B

4. Quick Check:

   Struct = grouped data [OK]

Hint: Structs bundle related info like a real-world object [OK]

Common Mistakes:

• Confusing structs with functions
• Thinking structs declare packages
• Mixing structs with imports

2. Which of the following is the correct way to define a struct named Person with fields Name (string) and Age (int)?

easy

A. type Person { Name string; Age int }

B. struct Person { Name string; Age int }

C. type Person struct { Name string; Age int }

D. Person struct { Name string; Age int }

Solution

1. Step 1: Recall Go struct syntax

   In Go, structs are defined using the syntax: type Name struct { fields }.

2. Step 2: Check each option

   type Person struct { Name string; Age int } matches the correct syntax. Options A, B, and C miss the 'type' keyword or 'struct' keyword placement.

3. Final Answer:

   type Person struct { Name string; Age int } -> Option C

4. Quick Check:

   Correct struct syntax = type Person struct { Name string; Age int } [OK]

Hint: Remember: 'type Name struct { fields }' [OK]

Common Mistakes:

• Omitting 'type' keyword
• Missing 'struct' keyword
• Incorrect order of keywords

3. What will be the output of the following Go code?

package main
import "fmt"
type Car struct {
  Brand string
  Year  int
}
func main() {
  c := Car{Brand: "Toyota", Year: 2020}
  fmt.Println(c.Brand)
}

medium

A. Car{Brand: "Toyota", Year: 2020}

B. 2020

C. Compilation error

D. Toyota

Solution

1. Step 1: Understand struct initialization

   The struct Car is initialized with Brand="Toyota" and Year=2020.

2. Step 2: Identify printed field

   The code prints c.Brand, which is "Toyota".

3. Final Answer:

   Toyota -> Option D

4. Quick Check:

   Print c.Brand = Toyota [OK]

Hint: Print field with dot: variable.FieldName [OK]

Common Mistakes:

• Printing the whole struct instead of a field
• Confusing field names
• Expecting Year instead of Brand

4. Identify the error in the following Go code snippet:

package main
import "fmt"
type Book struct {
  Title string
  Pages int
}

func main() {
  b := Book{Title: "Go Guide", Pages: "300"}
  fmt.Println(b)
}

medium

A. Pages field has wrong type, should be int not string

B. Missing import for fmt package

C. Struct fields must be lowercase

D. Cannot print struct directly

Solution

1. Step 1: Check field types in struct initialization

   The Pages field is declared as int but assigned a string "300" which is a type mismatch.

2. Step 2: Verify other possible errors

   Import for fmt package is present. Struct fields can be uppercase. Printing structs directly is allowed.

3. Final Answer:

   Pages field has wrong type, should be int not string -> Option A

4. Quick Check:

   Type mismatch in struct field assignment [OK]

Hint: Match field types exactly when initializing structs [OK]

Common Mistakes:

• Assigning string to int field
• Expecting compilation error from printing struct
• Thinking struct fields must be lowercase

5. Given the struct definition:

type Employee struct {
  Name   string
  Salary int
  Active bool
}

Which of the following code snippets correctly creates a slice of Employee structs with only active employees having salary above 50000?

hard

A. emps := []Employee{{"Alice", 60000, true}, {"Bob", 45000, true}, {"Carol", 70000, false}}
var filtered []Employee
for _, e := range emps {
if e.Active && e.Salary > 50000 {
filtered = append(filtered, e)
}
}

B. emps := []Employee{{"Alice", 60000, true}, {"Bob", 45000, true}, {"Carol", 70000, false}}
filtered := []Employee{}
for i := range emps {
if emps[i].Salary > 50000 {
filtered = append(filtered, emps[i])
}
}

C. emps := []Employee{{"Alice", 60000, true}, {"Bob", 45000, true}, {"Carol", 70000, false}}
filtered := []Employee{}
for _, e := range emps {
if e.Active || e.Salary > 50000 {
filtered = append(filtered, e)
}
}

D. emps := []Employee{{"Alice", 60000, true}, {"Bob", 45000, true}, {"Carol", 70000, false}}
filtered := []Employee{}
for _, e := range emps {
if e.Active && e.Salary < 50000 {
filtered = append(filtered, e)
}
}

Solution

1. Step 1: Understand filtering criteria

   We want only employees who are active and have salary above 50000.

2. Step 2: Check each code snippet logic

   emps := []Employee{{"Alice", 60000, true}, {"Bob", 45000, true}, {"Carol", 70000, false}} var filtered []Employee for _, e := range emps { if e.Active && e.Salary > 50000 { filtered = append(filtered, e) } } uses 'e.Active && e.Salary > 50000' which matches criteria. emps := []Employee{{"Alice", 60000, true}, {"Bob", 45000, true}, {"Carol", 70000, false}} filtered := []Employee{} for i := range emps { if emps[i].Salary > 50000 { filtered = append(filtered, emps[i]) } } misses active check. emps := []Employee{{"Alice", 60000, true}, {"Bob", 45000, true}, {"Carol", 70000, false}} filtered := []Employee{} for _, e := range emps { if e.Active || e.Salary > 50000 { filtered = append(filtered, e) } } uses '||' which includes inactive employees. emps := []Employee{{"Alice", 60000, true}, {"Bob", 45000, true}, {"Carol", 70000, false}} filtered := []Employee{} for _, e := range emps { if e.Active && e.Salary < 50000 { filtered = append(filtered, e) } } uses salary < 50000 which is wrong.

3. Final Answer:

   The code that filters using e.Active && e.Salary > 50000 -> Option A

4. Quick Check:

   Filter with '&&' for both conditions [OK]

Hint: Use && to combine multiple conditions correctly [OK]

Common Mistakes:

• Using || instead of && for filtering
• Ignoring the Active field in filter
• Using wrong salary comparison operator