Iterating over arrays in Go - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of range loop with index and value

What is the output of this Go program?

Go

package main
import "fmt"
func main() {
  arr := [3]int{10, 20, 30}
  for i, v := range arr {
    fmt.Printf("%d:%d ", i, v)
  }
}

A 0:10 1:20 2:30

B 10:0 20:1 30:2

C 0 10 1 20 2 30

D 10 20 30

💡 Hint

Remember that range returns index and value in that order.

The range keyword returns two values: the index and the element at that index. The fmt.Printf prints them as index:value pairs separated by spaces.

❓ Predict Output

intermediate

Output when ignoring index in range loop

What will this Go program print?

Go

package main
import "fmt"
func main() {
  arr := [4]string{"a", "b", "c", "d"}
  for _, v := range arr {
    fmt.Print(v)
  }
}

A abcd

B 0123

C a b c d

D [a b c d]

💡 Hint

The underscore _ ignores the index, so only values are printed.

The loop ignores the index and prints each string value directly without spaces or newlines.

🔧 Debug

advanced

Identify the error in this array iteration

What error will this Go code produce when compiled?

Go

package main
import "fmt"
func main() {
  arr := [3]int{1, 2, 3}
  for i, v := range arr {
    fmt.Println(i + v)
  }
  fmt.Println(arr[3])
}

A compile error: invalid array index 3 (out of bounds)

B no error, prints 1 3 5

C prints 1 3 5 and then 0

D runtime panic: index out of range

💡 Hint

Check the array length and the index used after the loop.

The array has length 3, so valid indices are 0,1,2. Accessing arr[3] causes a runtime panic: index out of range.

❓ Predict Output

advanced

Output of modifying array elements inside range loop

What is the output of this Go program?

Go

package main
import "fmt"
func main() {
  arr := [3]int{1, 2, 3}
  for i, v := range arr {
    v = v * 2
    fmt.Print(v, " ")
  }
  fmt.Println()
  fmt.Println(arr)
}

A

1 2 3 
[1 2 3]

B

2 4 6 
[2 4 6]

C

2 4 6 
[1 2 3]

D compile error

💡 Hint

Remember that v is a copy of the element, not the element itself.

The range loop copies each element into v. Modifying v does not change the original array. So printing v shows doubled values, but the array remains unchanged.

🧠 Conceptual

expert

Number of iterations in nested array loops

Given this Go code, how many times will the innermost print statement execute?

Go

package main
import "fmt"
func main() {
  arr := [2][3]int{{1,2,3},{4,5,6}}
  count := 0
  for _, inner := range arr {
    for i := range inner {
      fmt.Println(inner[i])
      count++
    }
  }
  fmt.Println("Total:", count)
}

A 5

B 6

C 3

D 2

💡 Hint

Count elements in each inner array and multiply by number of outer arrays.

The outer array has 2 elements, each inner array has 3 elements. The inner loop runs 3 times per outer iteration, total 2*3=6 times.

Practice

1. What does the range keyword do when used with an array in Go?

easy

A. It creates a new array with copied elements.

B. It changes the size of the array during iteration.

C. It lets you loop over each element and get its index and value.

D. It sorts the array elements automatically.

Solution

1. Step 1: Understand the role of range in loops

   The range keyword in Go is used to iterate over elements in arrays, slices, maps, and more, providing both the index and the value of each element.

2. Step 2: Identify what range does with arrays

   When used with an array, range returns the index and the value of each element one by one, allowing you to process or read each item.

3. Final Answer:

   It lets you loop over each element and get its index and value. -> Option C

4. Quick Check:

   range loops with index and value [OK]

Hint: Remember: range gives index and value for each element [OK]

Common Mistakes:

• Thinking range changes array size
• Assuming range sorts elements
• Believing range copies the array

2. Which of the following is the correct syntax to iterate over an array named nums in Go?

easy

A. for i, v := range nums { fmt.Println(i, v) }

B. for i in nums { fmt.Println(i) }

C. foreach i, v in nums { fmt.Println(i, v) }

D. for (i, v) in nums { fmt.Println(i, v) }

Solution

1. Step 1: Recall Go's for-range loop syntax

   In Go, the correct way to iterate over arrays or slices is using for index, value := range array syntax.

2. Step 2: Match the syntax with the options

   for i, v := range nums { fmt.Println(i, v) } uses for i, v := range nums, which is the correct Go syntax. Other options use invalid or non-Go syntax.

3. Final Answer:

   for i, v := range nums { fmt.Println(i, v) } -> Option A

4. Quick Check:

   Go uses for i, v := range array [OK]

Hint: Use 'for i, v := range array' for Go array loops [OK]

Common Mistakes:

• Using 'for i in nums' like Python
• Using 'foreach' which is not Go syntax
• Using parentheses in for loop header

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

nums := [3]int{10, 20, 30}
for i, v := range nums {
    fmt.Println(i, v)
}

medium

A. 0 0
1 1
2 2

B. 10 0
20 1
30 2

C. Syntax error

D. 0 10
1 20
2 30

Solution

1. Step 1: Understand the array and loop

   The array nums has 3 elements: 10 at index 0, 20 at index 1, and 30 at index 2. The loop uses range to get index and value.

2. Step 2: Trace the loop output

   Each iteration prints the index and the value: first line prints "0 10", second "1 20", third "2 30".

3. Final Answer:

   0 10 1 20 2 30 -> Option D

4. Quick Check:

   Index and value printed in order [OK]

Hint: range returns index then value in order [OK]

Common Mistakes:

• Swapping index and value in print
• Expecting values as indexes
• Thinking code has syntax error

4. Identify the error in this Go code snippet:

nums := [3]int{1, 2, 3}
for i, v = range nums {
    fmt.Println(i, v)
}

medium

A. Missing colon in 'for i, v = range nums' should be ':='

B. Using '=' instead of '==' in loop condition

C. Array declaration syntax is wrong

D. fmt.Println cannot print two variables

Solution

1. Step 1: Check the for loop variable declaration

   In Go, when declaring variables in a for-range loop, you must use ':=' to declare and assign variables, not '='.

2. Step 2: Identify the syntax error

   The code uses '=' instead of ':=' in for i, v = range nums, which causes a syntax error.

3. Final Answer:

   Missing colon in 'for i, v = range nums' should be ':=' -> Option A

4. Quick Check:

   Use ':=' to declare variables in for-range loops [OK]

Hint: Use ':=' not '=' in for-range variable declaration [OK]

Common Mistakes:

• Using '=' instead of ':=' in for loop
• Confusing assignment with comparison
• Incorrect array syntax blamed

5. Given an array nums := [5]int{1, 2, 3, 4, 5}, how can you create a new slice containing only the even numbers using iteration?

hard

A. Use a for loop to change the array size and remove odd numbers.

B. Use a for-range loop to check each value and append even numbers to a slice.

C. Use range to multiply all numbers by 2 and store in a new array.

D. Use a for loop with index only and ignore values.

Solution

1. Step 1: Understand the goal and array

   The goal is to create a new slice with only even numbers from the array nums. Arrays have fixed size, so we cannot remove elements from it.

2. Step 2: Use for-range loop and append even values

   We iterate over nums using for _, v := range nums, check if v % 2 == 0, and append those values to a new slice.

3. Final Answer:

   Use a for-range loop to check each value and append even numbers to a slice. -> Option B

4. Quick Check:

   Filter with for-range and append even numbers [OK]

Hint: Filter with for-range and append to slice [OK]

Common Mistakes:

• Trying to change array size directly
• Ignoring values in loop
• Multiplying instead of filtering