String Basics and Memory Representation in DSA Java - Build from Scratch

String Basics and Memory Representation

📖 Scenario: You are learning how strings work in Java. Strings are sequences of characters used to store words or sentences. Understanding how to create and compare strings is important for many programs.

🎯 Goal: You will create strings, compare them using == and equals(), and print the results to see how Java handles string memory and equality.

📋 What You'll Learn

Create string variables with exact values

Use == operator to compare string references

Use equals() method to compare string contents

Print the comparison results exactly as specified

💡 Why This Matters

🌍 Real World

Understanding string memory and comparison helps avoid bugs in programs that check user input, passwords, or text data.

💼 Career

Many software jobs require knowledge of how strings work in memory to write efficient and correct code, especially in Java backend development.

1

Create two string variables

Create two string variables called str1 and str2. Set str1 to "hello" and str2 to "hello".

DSA Java

# Your code here

Hint

Use double quotes to create string literals in Java.

2

Create a new string object

Add a new string variable called str3 and set it to a new String object with the value "hello" using new String("hello").

DSA Java

public class Main {
    public static void main(String[] args) {
        String str1 = "hello";
        String str2 = "hello";
        // Create str3 as a new String object with value "hello"
        
    }
}

Hint

Use new String("hello") to create a new string object.

3

Compare strings using == operator

Add three boolean variables called compare1, compare2, and compare3. Set compare1 to the result of str1 == str2, compare2 to str1 == str3, and compare3 to str2 == str3.

DSA Java

public class Main {
    public static void main(String[] args) {
        String str1 = "hello";
        String str2 = "hello";
        String str3 = new String("hello");
        // Compare strings using == operator
        
    }
}

Hint

Use == to compare if two string variables point to the same object.

4

Print comparison results using equals() and ==

Print the values of compare1, compare2, and compare3 each on a new line. Then print the results of str1.equals(str3) and str2.equals(str3) each on a new line.

DSA Java

public class Main {
    public static void main(String[] args) {
        String str1 = "hello";
        String str2 = "hello";
        String str3 = new String("hello");
        boolean compare1 = str1 == str2;
        boolean compare2 = str1 == str3;
        boolean compare3 = str2 == str3;
        // Print compare1, compare2, compare3
        // Print str1.equals(str3) and str2.equals(str3)
    }
}

Hint

Use System.out.println() to print each value on its own line.

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