Array initialization in Go - Step-by-Step Execution

Concept Flow - Array initialization

Declare array variable

↓

Specify array size

↓

Assign values to elements

↓

Array ready to use

This flow shows how an array is declared with a fixed size and then each element is assigned a value before use.

Execution Sample

Go

package main

import "fmt"

func main() {
    var arr [3]int
    arr[0] = 10
    arr[1] = 20
    arr[2] = 30
    fmt.Println(arr)
}

This code declares an integer array of size 3, assigns values to each element, and prints the array.

Execution Table

Step	Action	Array State	Output
1	Declare array 'arr' of size 3 with default values	[0 0 0]
2	Assign arr[0] = 10	[10 0 0]
3	Assign arr[1] = 20	[10 20 0]
4	Assign arr[2] = 30	[10 20 30]
5	Print arr	[10 20 30]	[10 20 30]

💡 All elements assigned, array printed, program ends.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	Final
arr	[0 0 0]	[10 0 0]	[10 20 0]	[10 20 30]	[10 20 30]

Key Moments - 3 Insights

Why does the array start with zeros before assignment?

In Go, arrays are initialized with zero values for their type automatically, as shown in step 1 of the execution_table.

Can we assign values beyond the declared size of the array?

No, the array size is fixed at declaration. Trying to assign beyond index 2 (for size 3) will cause a compile-time error.

What happens if we print the array before assigning any values?

The array will print with all zero values, as seen in step 1 where arr is [0 0 0].

Visual Quiz - 3 Questions

Look at the execution_table, what is the array state after step 3?

A [10 20 0]

B [10 0 0]

C [0 0 0]

D [10 20 30]

💡 Hint

Check the 'Array State' column in row with Step 3 in execution_table.

At which step is the array printed?

A Step 2

B Step 4

C Step 5

D Step 1

💡 Hint

Look for the 'Print arr' action in execution_table.

If we remove the assignment at step 4, what would be the final array state?

A [10 20 30]

B [10 20 0]

C [0 0 0]

D [10 0 0]

💡 Hint

Refer to variable_tracker and see the effect of skipping step 4 assignment.

Concept Snapshot

Array initialization in Go:
- Declare with fixed size: var arr [3]int
- Elements start with zero values
- Assign values by index: arr[0] = 10
- Size cannot change after declaration
- Use fmt.Println(arr) to print entire array

Full Transcript

This example shows how to create an array in Go. First, we declare an array named arr with size 3. Go automatically fills it with zeros. Then, we assign values 10, 20, and 30 to each element by their index. Finally, we print the array, which outputs [10 20 30]. Arrays have fixed size, so you cannot add more elements than declared.

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