Constants in Go - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of constant iota usage

What is the output of this Go program?

Go

package main
import "fmt"
const (
  a = iota
  b
  c = 5
  d
  e = iota
)
func main() {
  fmt.Println(a, b, c, d, e)
}

A 0 1 5 5 4

B 0 1 5 5 5

C 0 1 5 6 4

D 0 1 5 6 5

💡 Hint

Remember that iota increments on each line, but explicit assignments reset it.

iota starts at 0 for 'a', increments by 1 for 'b'. 'c' is explicitly 5, so 'd' repeats 5. 'e' resumes iota counting from the line number, which is 4.

❓ Predict Output

intermediate

Constant expression evaluation

What will this Go program print?

Go

package main
import "fmt"
const (
  x = 3
  y = x * 2
  z = y + 4
)
func main() {
  fmt.Println(x, y, z)
}

A 3 6 10

B 3 6 7

C 3 5 9

D 3 6 12

💡 Hint

Constants can be expressions using other constants.

x is 3, y is 3*2=6, z is 6+4=10.

❓ Predict Output

advanced

Constant block with mixed types

What is the output of this Go program?

Go

package main
import "fmt"
const (
  a = 1
  b
  c = "hello"
  d
  e = 3.14
  f
)
func main() {
  fmt.Printf("%v %v %v %v %v %v", a, b, c, d, e, f)
}

A 1 2 hello hello 3.14 3.15

B 1 1 hello hello 3.14 3.14

C 1 1 hello hello 3.14 0

D 1 1 hello hello 3.14 <nil>

💡 Hint

Constants repeat the previous value if no new value is given.

b repeats a's value 1, d repeats c's value "hello", f repeats e's value 3.14.

❓ Predict Output

advanced

Using iota with bit shifting

What does this Go program print?

Go

package main
import "fmt"
const (
  _ = 1 << (iota * 2)
  A
  B
  C
)
func main() {
  fmt.Println(A, B, C)
}

A 2 8 32

B 1 4 16

C 4 16 64

D 0 4 16

💡 Hint

iota starts at 0, but the first constant is ignored with underscore.

The first line sets _ = 1 << (0*2) = 1, ignored. Then A = 1 << (1*2) = 4, B = 1 << (2*2) = 16, C = 1 << (3*2) = 64.

🧠 Conceptual

expert

Why are constants preferred over variables for fixed values?

Which is the best reason to use constants instead of variables for fixed values in Go?

A Constants allow dynamic typing which variables do not.

B Constants use less memory than variables because they are stored on the stack.

C Constants can be changed during program execution for flexibility.

D Constants improve code readability and prevent accidental changes at runtime.

💡 Hint

Think about safety and clarity in code.

Constants cannot be changed after compilation, which helps avoid bugs and makes code easier to understand.

Practice

1. What is the purpose of using const in Go programming?

easy

A. To declare a value that cannot be changed

B. To declare a variable that can change

C. To create a function

D. To start a loop

Solution

1. Step 1: Understand the role of const

   The const keyword is used to declare constants, which are fixed values that cannot be changed after being set.

2. Step 2: Compare options with the definition

   To declare a value that cannot be changed correctly states that const declares a value that cannot be changed, while other options describe unrelated concepts.

3. Final Answer:

   To declare a value that cannot be changed -> Option A

4. Quick Check:

   Constants = fixed values [OK]

Hint: Constants hold fixed values, variables do not change [OK]

Common Mistakes:

• Confusing constants with variables
• Thinking const declares functions
• Assuming const starts loops

2. Which of the following is the correct way to declare a constant named Pi with value 3.14 in Go?

easy

A. const Pi = 3.14

B. let Pi = 3.14

C. Pi := 3.14

D. var Pi = 3.14

Solution

1. Step 1: Recall Go syntax for constants

   In Go, constants are declared using the const keyword followed by the name and value.

2. Step 2: Identify the correct syntax

   const Pi = 3.14 uses const Pi = 3.14, which is the correct syntax. Options A and C declare variables, and B uses invalid Go syntax.

3. Final Answer:

   const Pi = 3.14 -> Option A

4. Quick Check:

   const keyword = correct declaration [OK]

Hint: Use const keyword for fixed values, not var or := [OK]

Common Mistakes:

• Using var or := instead of const
• Using let which is not Go syntax
• Missing the const keyword

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

package main
import "fmt"
const greeting = "Hello"
func main() {
    fmt.Println(greeting)
}

medium

A. greeting

B. Compilation error

C. Hello\n

D. Hello

Solution

1. Step 1: Understand constant usage in the program

   The constant greeting is declared with value "Hello" and printed in main.

2. Step 2: Predict the output of fmt.Println

   fmt.Println(greeting) prints the string value of the constant, which is "Hello" without quotes.

3. Final Answer:

   Hello -> Option D

4. Quick Check:

   Printing constant value = Hello [OK]

Hint: Print constant name to get its fixed value output [OK]

Common Mistakes:

• Expecting the name 'greeting' to print
• Thinking it prints with quotes
• Assuming a compile error

4. Identify the error in this Go code snippet:

package main
import "fmt"
const x = 10
func main() {
    x = 20
    fmt.Println(x)
}

medium

A. Variable 'x' not declared

B. Cannot assign to constant 'x' after declaration

C. Missing import of fmt package

D. No error, code runs fine

Solution

1. Step 1: Check constant reassignment

   The code declares x as a constant with value 10, then tries to assign 20 to it, which is not allowed.

2. Step 2: Verify other possible errors

   There are no other compilation errors; the only issue is reassignment of the constant.

3. Final Answer:

   Cannot assign to constant 'x' after declaration -> Option B

4. Quick Check:

   Constants cannot be reassigned [OK]

Hint: Constants cannot be changed after declaration [OK]

Common Mistakes:

• Trying to assign new value to a constant
• Ignoring missing imports
• Confusing constants with variables

5. Given the following Go code, what will be the output?

package main
import "fmt"
const (
    A = iota
    B
    C = 5
    D
)
func main() {
    fmt.Println(A, B, C, D)
}

hard

A. 0 1 2 3

B. 0 1 5 5

C. 0 1 5 3

D. Compilation error due to iota usage

Solution

1. Step 1: Understand iota behavior in constant blocks

   iota starts at 0 and increments by 1 for each line unless overridden.

2. Step 2: Analyze each constant value

   A = iota (0), iota increments to 1; B = iota (1), iota to 2; C = 5 (explicit), iota to 3; D = iota (3).

3. Step 3: Confirm output

   Printing A, B, C, D outputs: 0 1 5 3

4. Final Answer:

   0 1 5 3 -> Option C

5. Quick Check:

   iota increments except when overridden [OK]

Hint: iota increments unless constant value is explicitly set [OK]

Common Mistakes:

• Assuming D equals C's value (5)
• Thinking iota resets after explicit value
• Expecting compilation error