Constants in Go - Step-by-Step Execution

Concept Flow - Constants

Declare constant with const keyword

↓

Assign fixed value

↓

Use constant in code

↓

Value cannot change

↓

Program runs

Constants are fixed values set once and used throughout the program without change.

Execution Sample

Go

package main
import "fmt"
const Pi = 3.14
func main() {
    fmt.Println(Pi)
}

This code declares a constant Pi and prints its value.

Execution Table

Step	Action	Evaluation	Result
1	Declare constant Pi	const Pi = 3.14	Pi is set to 3.14
2	Enter main function	func main()	Ready to execute main
3	Print Pi	fmt.Println(Pi)	Outputs 3.14
4	End program	No further code	Program terminates

💡 Program ends after printing the constant Pi value

Variable Tracker

Variable	Start	After 1	After 2	Final
Pi	undefined	3.14	3.14	3.14

Key Moments - 2 Insights

Why can't we change the value of Pi after declaring it?

Because Pi is declared as a constant (see execution_table step 1), its value is fixed and cannot be reassigned.

What happens if we try to assign a new value to Pi?

The Go compiler will give an error since constants cannot be changed after declaration, preventing reassignment.

Visual Quiz - 3 Questions

Look at the execution table, what is the value of Pi at step 3?

A 3.14

B undefined

C 0

D Error

💡 Hint

Check the 'Evaluation' and 'Result' columns at step 3 in the execution_table.

At which step does the program print the constant value?

A Step 1

B Step 2

C Step 3

D Step 4

💡 Hint

Look for the action 'Print Pi' in the execution_table.

If we tried to assign Pi = 3.1415 after declaration, what would happen?

A Pi changes to 3.1415

B Compiler error

C Program prints 3.1415

D Program ignores the assignment

💡 Hint

Constants cannot be reassigned as explained in key_moments and execution_table step 1.

Concept Snapshot

Constants in Go:
- Declared with 'const' keyword
- Assigned a fixed value at declaration
- Value cannot be changed later
- Used like variables but immutable
- Helps prevent accidental value changes

Full Transcript

This visual execution shows how constants work in Go. First, a constant Pi is declared with the value 3.14. This value is fixed and cannot be changed later. The main function prints the value of Pi. The program ends after printing. Trying to change Pi would cause a compiler error because constants are immutable.

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