Message serialization in Spring Boot - Step-by-Step Execution

Concept Flow - Message serialization

Create Java Object

↓

Serialize Object to JSON

↓

Send JSON over Network

↓

Receive JSON

↓

Deserialize JSON to Java Object

↓

Use Java Object in Application

This flow shows how a Java object is converted to JSON for sending, then back to a Java object when received.

Execution Sample

Spring Boot

import com.fasterxml.jackson.databind.ObjectMapper;

public class User {
  private String name;
  private int age;

  public User() {}

  public User(String name, int age) {
    this.name = name;
    this.age = age;
  }

  public String getName() {
    return name;
  }

  public void setName(String name) {
    this.name = name;
  }

  public int getAge() {
    return age;
  }

  public void setAge(int age) {
    this.age = age;
  }
}

User user = new User("Alice", 30);
String json = new ObjectMapper().writeValueAsString(user);

This code creates a User object and converts it to a JSON string using Jackson's ObjectMapper.

Execution Table

Step	Action	Input	Output	Notes
1	Create User object	name="Alice", age=30	User{name='Alice', age=30}	Java object created in memory
2	Serialize to JSON	User object	{"name":"Alice","age":30}	ObjectMapper converts object to JSON string
3	Send JSON	{"name":"Alice","age":30}	JSON string sent over network	JSON is text format suitable for transfer
4	Receive JSON	JSON string	JSON string received	Data arrives as JSON text
5	Deserialize JSON	{"name":"Alice","age":30}	User{name='Alice', age=30}	ObjectMapper converts JSON back to Java object
6	Use Java object	User object	Access user.name and user.age	Application uses deserialized object
7	End	N/A	N/A	Process complete

💡 Serialization and deserialization complete, object ready for use

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 5	Final
user	null	User{name='Alice', age=30}	User{name='Alice', age=30}	User{name='Alice', age=30}	User{name='Alice', age=30}
json	null	null	{"name":"Alice","age":30}	{"name":"Alice","age":30}	{"name":"Alice","age":30}

Key Moments - 2 Insights

Why do we convert a Java object to JSON before sending it?

Because JSON is a text format that can be sent over networks easily, unlike Java objects which only exist in memory (see execution_table step 2 and 3).

What happens if the JSON structure does not match the Java class during deserialization?

Deserialization will fail or produce incorrect objects because ObjectMapper expects JSON fields to match Java class fields (refer to execution_table step 5).

Visual Quiz - 3 Questions

Look at the execution table, what is the value of 'json' after step 2?

A User{name='Alice', age=30}

B {"name":"Alice","age":30}

C null

D JSON object

💡 Hint

Check the 'Output' column at step 2 in the execution_table.

At which step does the Java object get recreated from JSON?

A Step 5

B Step 3

C Step 2

D Step 4

💡 Hint

Look for 'Deserialize JSON' action in the execution_table.

If the 'user' object had an extra field not in JSON, what would happen during deserialization?

A Deserialization ignores the extra field and succeeds

B Deserialization fails with error

C Extra field gets default value

D JSON adds the extra field automatically

💡 Hint

Think about how ObjectMapper handles missing JSON fields during deserialization (see key_moments).

Concept Snapshot

Message serialization in Spring Boot:
- Convert Java object to JSON string using ObjectMapper
- Send JSON over network as text
- Receive JSON and convert back to Java object
- JSON must match Java class structure
- Enables data exchange between systems

Full Transcript

Message serialization means turning a Java object into a JSON string so it can be sent over a network. In Spring Boot, we use ObjectMapper to do this. First, we create a Java object like a User with name and age. Then ObjectMapper converts it to JSON text. This JSON is sent over the network. When received, ObjectMapper reads the JSON and creates a new Java object with the same data. This process allows different systems to share data easily. The JSON structure must match the Java class fields to work correctly.