String Basics and Memory Representation in DSA Java - Execution Trace

Concept Flow - String Basics and Memory Representation

Create String Object

↓

Store Characters in Array

↓

Reference String Object

↓

Use String in Program

↓

Immutable: No Change Allowed

↓

New String Created if Modified

This flow shows how a string is created, stored as characters in memory, referenced, used, and how immutability means changes create new strings.

Execution Sample

DSA Java

String s = "hello";
System.out.println(s);
s = s + " world";
System.out.println(s);

Creates a string 'hello', prints it, then creates a new string by adding ' world' and prints the new string.

Execution Table

Step	Operation	String Object(s)	Memory Representation	Visual State
1	Create String s = "hello"	s -> "hello"	[h][e][l][l][o]	"hello"
2	Print s	s -> "hello"	[h][e][l][l][o]	Output: hello
3	Concatenate s + " world"	s -> "hello world" (new)	[h][e][l][l][o][ ][w][o][r][l][d]	"hello world"
4	Print s	s -> "hello world"	[h][e][l][l][o][ ][w][o][r][l][d]	Output: hello world
5	End	No change	No change	Program ends

💡 Program ends after printing the concatenated string; original string 'hello' remains unchanged due to immutability.

Variable Tracker

Variable	Start	After Step 1	After Step 3	Final
s	null	"hello"	"hello world"	"hello world"

Key Moments - 3 Insights

Why does concatenating strings create a new string instead of changing the original?

Strings are immutable in Java, so operations like concatenation create new string objects rather than modifying existing ones, as shown in execution_table step 3.

What does the memory representation of a string look like?

Each string is stored as a sequence of characters in an array, as shown in the Memory Representation column in execution_table steps 1 and 3.

Why does printing the string not change its memory?

Printing only reads the string's characters without modifying them, so the memory representation stays the same, as seen in execution_table steps 2 and 4.

Visual Quiz - 3 Questions

Look at the execution_table at step 3, what happens to variable s?

A s points to a new string "hello world"

B s points to the original string "hello"

C s is set to null

D s is unchanged

💡 Hint

Check the String Object(s) column at step 3 in execution_table.

At which step does the program output the string "hello"?

A Step 1

B Step 3

C Step 2

D Step 4

💡 Hint

Look at the Visual State column for output in execution_table.

If strings were mutable, how would the memory representation change after concatenation at step 3?

A A new array would still be created

B The original array would be modified to add new characters

C The variable s would become null

D The string would be deleted

💡 Hint

Think about how immutability affects memory shown in execution_table step 3.

Concept Snapshot

String Basics and Memory Representation:
- Strings are objects storing characters in arrays.
- Strings are immutable: cannot change after creation.
- Concatenation creates new string objects.
- Variables reference string objects in memory.
- Printing reads string without changing it.

Full Transcript

This visual trace shows how strings in Java are created and stored as arrays of characters. When we create a string like "hello", it is stored in memory as characters in sequence. The variable s points to this string object. When we print s, the characters are read and output without changing the string. When we concatenate s with another string, a new string object is created with combined characters, and s points to this new object. The original string remains unchanged because strings are immutable in Java. This immutability means any modification results in a new string object rather than changing the existing one.

Practice

1. What is the correct way to create a string literal in Java?

easy

A. String s = 'Hello';

B. String s = new String();

C. String s = "Hello";

D. String s = String();

Solution

1. Step 1: Understand string literal syntax

   In Java, string literals are enclosed in double quotes, so "Hello" is correct.

2. Step 2: Check each option

   String s = "Hello"; uses double quotes correctly. String s = new String(); creates an empty string object but not a literal. String s = 'Hello'; uses single quotes which are for characters, not strings. String s = String(); is invalid syntax.

3. Final Answer:

   String s = "Hello"; -> Option C

4. Quick Check:

   String literal uses double quotes = A [OK]

Hint: Use double quotes for string literals in Java [OK]

Common Mistakes:

• Using single quotes for strings
• Trying to call String() like a function
• Confusing new String() with literal

2. Which of the following is the correct syntax to compare two strings s1 and s2 for content equality in Java?

easy

A. if (s1.equals(s2))

B. if (s1 == s2)

C. if (s1 = s2)

D. if (s1.compare(s2))

Solution

1. Step 1: Understand string comparison in Java

   Using == compares references (memory addresses), not content. To compare content, use the equals() method.

2. Step 2: Evaluate each option

   if (s1 == s2) compares references, not content. if (s1.equals(s2)) correctly uses equals() method. if (s1 = s2) uses assignment (=) which is invalid in condition. if (s1.compare(s2)) uses a non-existent method compare().

3. Final Answer:

   if (s1.equals(s2)) -> Option A

4. Quick Check:

   Use equals() for string content comparison = B [OK]

Hint: Use equals() method to compare string contents [OK]

Common Mistakes:

• Using == to compare strings
• Using assignment operator in if condition
• Calling non-existent methods

3. What will be the output of the following Java code?

String s1 = "hello";
String s2 = "hello";
String s3 = new String("hello");
System.out.println(s1 == s2);
System.out.println(s1 == s3);
System.out.println(s1.equals(s3));

medium

A. true
false
true

B. false
false
true

C. true
true
true

D. false
true
false

Solution

1. Step 1: Understand string pool and new String()

   s1 and s2 are string literals and point to the same object in the string pool, so s1 == s2 is true. s3 is a new String object, so s1 == s3 is false because they reference different objects.

2. Step 2: Check equals() method

   s1.equals(s3) compares content and both have "hello", so it returns true.

3. Final Answer:

   true false true -> Option A

4. Quick Check:

   == compares references, equals() compares content = C [OK]

Hint: == checks memory, equals() checks string content [OK]

Common Mistakes:

• Assuming new String() shares reference
• Confusing == with equals()
• Ignoring string pool behavior

4. Identify the error in the following code snippet:

String s = "test";
if (s == "test") {
    System.out.println("Match");
} else {
    System.out.println("No Match");
}

medium

A. System.out.println syntax is incorrect

B. Missing semicolon after string declaration

C. String literals cannot be compared

D. Using == to compare strings can cause unexpected results

Solution

1. Step 1: Check string comparison method

   The code uses == to compare strings, which compares references, not content. This can cause unexpected results if strings are not from the pool.

2. Step 2: Verify other syntax

   There is no missing semicolon, string literals can be compared with equals(), and System.out.println syntax is correct.

3. Final Answer:

   Using == to compare strings can cause unexpected results -> Option D

4. Quick Check:

   Use equals() for string comparison, not == = A [OK]

Hint: Avoid == for strings; use equals() instead [OK]

Common Mistakes:

• Thinking == compares string content
• Ignoring string pool behavior
• Assuming syntax errors where none exist

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

String a = "Java";
String b = "Ja" + "va";
String c = new String("Java");
System.out.println(a == b);
System.out.println(a == c);
System.out.println(a.equals(c));

Explain why the outputs are as they are.

hard

A. false
true
false

B. true
false
true

C. true
true
true

D. false
false
true

Solution

1. Step 1: Analyze string concatenation with literals

   Strings "Ja" and "va" are literals concatenated at compile time, so b points to the same string pool object as a. Therefore, a == b is true.

2. Step 2: Understand new String() behavior

   c is a new String object created at runtime, so a == c is false because they reference different objects. However, a.equals(c) compares content and returns true.

3. Final Answer:

   true false true -> Option B

4. Quick Check:

   Compile-time concat shares pool, new String() creates new object = D [OK]

Hint: Compile-time concat shares pool; new String() creates new object [OK]

Common Mistakes:

• Assuming runtime concat shares pool
• Confusing == with equals()
• Ignoring compile-time optimization