Array initialization in Go

Introduction

Arrays hold a fixed number of items of the same type. Initializing an array means creating it and setting its starting values.

When you know the exact number of items you want to store.

When you want to store a list of fixed size, like days in a week or months in a year.

When you want to prepare a container to hold values before filling it later.

When you want to set default values for a group of items at once.

Syntax

Go

package main

import "fmt"

type ArrayExample struct {
    items [5]int
}

func main() {
    var example ArrayExample
    example.items = [5]int{1, 2, 3, 4, 5} // Initialize array with values
    fmt.Println(example.items)
}

Arrays in Go have a fixed size defined at compile time.

You can initialize arrays with values using curly braces {}.

Examples

This creates an empty array of size 3 with default zero values.

Go

var numbers [3]int
// Creates an array of 3 integers, all set to 0 by default

Here, the array is created with 7 days of the week as initial values.

Go

var days = [7]string{"Mon", "Tue", "Wed", "Thu", "Fri", "Sat", "Sun"}
// Creates and initializes an array of 7 strings

This shows an array with just one element.

Go

var singleElement = [1]int{42}
// Array with only one element

An array with zero length is allowed but holds no elements.

Go

var emptyArray [0]int
// Array with zero length

Sample Program

This program creates an array with 5 numbers, prints it, changes the first and last values, then prints it again.

Go

package main

import "fmt"

func main() {
    // Initialize an array of 5 integers with values
    var numbers = [5]int{10, 20, 30, 40, 50}

    fmt.Println("Array before change:", numbers)

    // Change the first and last elements
    numbers[0] = 100
    numbers[4] = 500

    fmt.Println("Array after change:", numbers)
}

Important Notes

Time complexity to access or change an element is O(1) because arrays allow direct access by index.

Space complexity is fixed and depends on the array size declared.

Common mistake: Trying to add more elements than the array size causes a compile-time error.

Use arrays when the size is fixed and known. Use slices when the size can change.

Summary

Arrays hold a fixed number of items of the same type.

You create arrays by specifying size and optionally initial values.

Access and change elements by their index quickly.

Practice

1. Which of the following correctly declares an array of 5 integers in Go?

easy

A. var arr int[5]

B. var arr [5]int

C. arr := []int{5}

D. arr := [5]int{1, 2, 3, 4}

Solution

1. Step 1: Understand Go array declaration syntax

   In Go, arrays are declared with the syntax: var name [size]type.

2. Step 2: Check each option

   A: var arr int[5] wrong syntax (type before size). B: arr := []int{5} is a slice, not an array. C: var arr [5]int correct syntax. D: arr := [5]int{1, 2, 3, 4} too few initializers (4 < 5), compilation error.

3. Final Answer:

   var arr [5]int -> Option B

4. Quick Check:

   Array declaration = var arr [5]int [OK]

Hint: Remember: var name [size]type declares an array [OK]

Common Mistakes:

• Confusing slices with arrays
• Using wrong syntax like int[5]
• Missing var keyword for array declaration

2. Which of these is the correct way to initialize an array of 3 strings with values "a", "b", "c" in Go?

easy

A. arr := [3]string{"a", "b", "c"}

B. arr := []string{"a", "b", "c"}

C. var arr [3]string = {"a", "b", "c"}

D. arr := [3]string{"a", "b", c}

Solution

1. Step 1: Identify correct array initialization syntax

   Arrays use [size]type{values} syntax. arr := [3]string{"a", "b", "c"} matches this exactly.

2. Step 2: Check other options for errors

   arr := []string{"a", "b", "c"} is a slice, not an array. var arr [3]string = {"a", "b", "c"} uses invalid initialization with = and braces. arr := [3]string{"a", "b", c} misses quotes around "c" causing syntax error.

3. Final Answer:

   arr := [3]string{"a", "b", "c"} -> Option A

4. Quick Check:

   Array init with values = [3]string{"a", "b", "c"} [OK]

Hint: Use [size]type{values} for array initialization [OK]

Common Mistakes:

• Using slice syntax instead of array
• Missing quotes around string elements
• Incorrect use of = with array literals

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

package main
import "fmt"
func main() {
  arr := [4]int{1, 2}
  fmt.Println(arr)
}

medium

A. [1 2 0 0]

B. [1 2]

C. [0 0 0 0]

D. Compilation error

Solution

1. Step 1: Understand partial array initialization

   When fewer elements are provided than the array size, Go fills remaining elements with zero values.

2. Step 2: Apply to given array

   Array size is 4, initialized with 1 and 2. Remaining two elements become 0 by default.

3. Final Answer:

   [1 2 0 0] -> Option A

4. Quick Check:

   Partial init fills rest with zero [OK]

Hint: Uninitialized array elements default to zero [OK]

Common Mistakes:

• Assuming array length matches initializer length
• Expecting error for fewer initial values
• Confusing slices with arrays

4. Identify the error in this Go array initialization:

var arr [3]int = [3]int{1, 2, 3, 4}

medium

A. No error, code is correct

B. Missing var keyword

C. Incorrect type declaration

D. Array size mismatch with initializer

Solution

1. Step 1: Check array size and initializer length

   Array declared with size 3 but initializer has 4 elements.

2. Step 2: Understand Go array rules

   Go requires initializer length to match array size exactly, else compilation error occurs.

3. Final Answer:

   Array size mismatch with initializer -> Option D

4. Quick Check:

   Initializer length must equal array size [OK]

Hint: Initializer length must match array size exactly [OK]

Common Mistakes:

• Ignoring extra elements in initializer
• Assuming Go auto-expands array size
• Confusing slices with arrays

5. You want to create an array of 5 integers where the first two elements are 10 and 20, and the rest are zeros. Which code snippet correctly achieves this and prints the array?

hard

A. arr := [5]int{10, 20, 30}
fmt.Println(arr)

B. arr := [3]int{10, 20}
fmt.Println(arr)

C. arr := [5]int{10, 20, 0, 0, 0}
fmt.Println(arr)

D. arr := []int{10, 20}
fmt.Println(arr)

Solution

1. Step 1: Understand array initialization with explicit zeros

   To ensure the last three elements are zeros explicitly, include them in the initializer.

2. Step 2: Check each option

   B: arr := [3]int{10, 20} fmt.Println(arr) wrong size 3, not 5. A: arr := [5]int{10, 20, 0, 0, 0} fmt.Println(arr) explicitly sets zeros. C: arr := [5]int{10, 20, 30} fmt.Println(arr) third element 30, not zero. D: arr := []int{10, 20} fmt.Println(arr) is a slice, not array.

3. Step 3: Choose the best explicit solution

   A clearly shows all elements matching the requirement exactly.

4. Final Answer:

   arr := [5]int{10, 20, 0, 0, 0} fmt.Println(arr) -> Option C

5. Quick Check:

   Explicit zeros in initializer = arr := [5]int{10, 20, 0, 0, 0} fmt.Println(arr) [OK]

Hint: Explicitly list zeros to avoid confusion [OK]

Common Mistakes:

• Assuming missing elements are always zero without clarity
• Using slices instead of arrays
• Adding unintended values in initializer