Array initialization in Go - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of array initialization with default values

What is the output of this Go program?

package main
import "fmt"
func main() {
  var arr [4]int
  fmt.Println(arr)
}

Go

package main
import "fmt"
func main() {
  var arr [4]int
  fmt.Println(arr)
}

A [0 1 2 3]

B [1 2 3 4]

C [0 0 0 0]

D [4 4 4 4]

💡 Hint

In Go, arrays of integers are initialized with zero values by default.

When you declare an array of integers without assigning values, Go fills it with zeros.

❓ Predict Output

intermediate

Output of array initialization with explicit values

What does this Go program print?

package main
import "fmt"
func main() {
  arr := [3]string{"go", "lang", "rocks"}
  fmt.Println(arr)
}

Go

package main
import "fmt"
func main() {
  arr := [3]string{"go", "lang", "rocks"}
  fmt.Println(arr)
}

A [go lang rocks rocks]

B ["go" "lang" "rocks"]

C [go lang]

D [go lang rocks]

💡 Hint

Arrays print their elements separated by spaces inside square brackets.

The array is initialized with three strings and prints exactly those elements.

❓ Predict Output

advanced

Output of partially initialized array

What is the output of this Go program?

package main
import "fmt"
func main() {
  arr := [5]int{1, 2}
  fmt.Println(arr)
}

Go

package main
import "fmt"
func main() {
  arr := [5]int{1, 2}
  fmt.Println(arr)
}

A [1 2 0 0 0]

B [1 2 3 4 5]

C [0 0 0 0 0]

D [1 2]

💡 Hint

Uninitialized elements in an array get zero values.

Only the first two elements are set; the rest default to zero.

❓ Predict Output

advanced

Output of array initialization with index keys

What does this Go program print?

package main
import "fmt"
func main() {
  arr := [4]int{1: 10, 3: 30}
  fmt.Println(arr)
}

Go

package main
import "fmt"
func main() {
  arr := [4]int{1: 10, 3: 30}
  fmt.Println(arr)
}

A [10 30 0 0]

B [0 10 0 30]

C [1 10 3 30]

D [10 0 30 0]

💡 Hint

Elements not explicitly set get zero values.

Only indexes 1 and 3 are set; others are zero.

🧠 Conceptual

expert

Length of array initialized with ellipsis

What is the length of the array declared in this Go code?

arr := [...]int{5, 10, 15, 20}

Go

package main
import "fmt"
func main() {
  arr := [...]int{5, 10, 15, 20}
  fmt.Println(len(arr))
}

A 4

B 3

C 5

D 0

💡 Hint

The ellipsis lets Go count the elements automatically.

The array length is the number of elements inside the braces.

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