Data class copy and destructuring in Kotlin - Step-by-Step Execution

Concept Flow - Data class copy and destructuring

Create data class instance

↓

Destructure instance into variables

↓

Use copy() to create modified instance

↓

Compare original and copied instances

↓

End

This flow shows creating a data class object, breaking it into variables, copying it with changes, and comparing both objects.

Execution Sample

Kotlin

data class User(val name: String, val age: Int)

fun main() {
  val user1 = User("Alice", 30)
  val (name, age) = user1
  val user2 = user1.copy(age = 31)
  println("$name is $age years old")
  println(user2)
}

Creates a User, destructures it, copies with a changed age, then prints values.

Execution Table

Step	Action	Evaluation	Result
1	Create user1	User("Alice", 30)	user1 = User(name=Alice, age=30)
2	Destructure user1	(name, age) = user1	name = "Alice", age = 30
3	Copy user1 with age=31	user1.copy(age=31)	user2 = User(name=Alice, age=31)
4	Print name and age	println("$name is $age years old")	Output: Alice is 30 years old
5	Print user2	println(user2)	Output: User(name=Alice, age=31)
6	End	No more code	Program ends

💡 Program ends after printing both original and copied user details.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	Final
user1	undefined	User(name=Alice, age=30)	User(name=Alice, age=30)	User(name=Alice, age=30)	User(name=Alice, age=30)
name	undefined	undefined	"Alice"	"Alice"	"Alice"
age	undefined	undefined	30	30	30
user2	undefined	undefined	undefined	User(name=Alice, age=31)	User(name=Alice, age=31)

Key Moments - 2 Insights

Why does the destructured 'age' variable stay 30 even after copying user1 with age 31?

Because destructuring happens before the copy. The variable 'age' holds the original value from user1 (30), not from the copied user2.

Does the copy() function change the original user1 object?

No, copy() creates a new object with the specified changes. user1 remains unchanged as shown in step 3 and variable_tracker.

Visual Quiz - 3 Questions

Look at the execution table at step 3. What is the value of user2?

A User(name=Alice, age=30)

B User(name=Alice, age=31)

C User(name=Bob, age=31)

D User(name=Bob, age=30)

💡 Hint

Check the 'Result' column at step 3 in the execution_table.

At which step does the variable 'name' get assigned the value "Alice"?

A Step 1

B Step 3

C Step 2

D Step 4

💡 Hint

Look at the 'Action' and 'Result' columns for when destructuring happens.

If we changed user1.copy(age = 35) instead of 31, what would user2's age be in the variable_tracker after step 3?

A 35

B 31

C 30

D Undefined

💡 Hint

The copy() function sets the new age value in user2 as shown in step 3.

Concept Snapshot

Data class copy and destructuring in Kotlin:
- Use 'val (a, b) = obj' to unpack properties.
- Use 'obj.copy(prop = newVal)' to create modified copies.
- Original object stays unchanged.
- Destructured variables keep original values.
- Useful for easy object cloning and unpacking.

Full Transcript

This example shows how Kotlin data classes let you easily copy objects and break them into variables. First, we create a User named Alice aged 30. Then we unpack her name and age into separate variables. Next, we make a copy of Alice but change her age to 31. When printing, the destructured age is still 30 because it was taken before copying. The copy creates a new User with the updated age, leaving the original unchanged. This helps manage data cleanly and simply.