Why strings are special in Java - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a string variable named greeting.

Java

String greeting = [1];

Drag options to blanks, or click blank then click option'

A"Hello"

BHello

C'Hello'

Dnew String

Attempts:

3 left

💡 Hint

Common Mistakes

Using single quotes instead of double quotes for strings

Not using quotes at all

Strings in Java are enclosed in double quotes. So, "Hello" is the correct way to assign a string literal.

2fill in blank

medium

Complete the code to compare two strings for equality.

Java

if (str1.[1](str2)) {
    System.out.println("Strings are equal");
}

Drag options to blanks, or click blank then click option'

AcompareTo

Bequals

C==

DequalsIgnoreCase

Attempts:

3 left

💡 Hint

Common Mistakes

Using == to compare strings

Using compareTo without checking the result

In Java, to check if two strings have the same content, use the equals() method.

3fill in blank

hard

Fix the error in the code to concatenate two strings correctly.

Java

String fullName = firstName [1] lastName;

Drag options to blanks, or click blank then click option'

Aadd()

Bconcat()

C&

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using & which is not valid for strings in Java

Trying to use add() which is not a string method

In Java, the plus sign (+) is used to join strings together.

4fill in blank

hard

Fill both blanks to create a string and get its length.

Java

String word = [1];
int length = word.[2]();

Drag options to blanks, or click blank then click option'

A"Java"

Blength

Clength()

Dsize()

Attempts:

3 left

💡 Hint

Common Mistakes

Using length without parentheses

Using size() which is for collections, not strings

Strings are created with double quotes. To get the number of characters, use the length() method.

5fill in blank

hard

Fill all three blanks to check if a string starts with a letter and convert it to uppercase.

Java

if (text.[1]("A")) {
    String upper = text.[2]();
    System.out.println(upper.[3]());
}

Drag options to blanks, or click blank then click option'

AstartsWith

BtoUpperCase

CtoLowerCase

Dtrim

Attempts:

3 left

💡 Hint

Common Mistakes

Using toLowerCase instead of toUpperCase

Using trim which removes spaces, not changes case

Use startsWith() to check the beginning of a string, toUpperCase() to convert to uppercase, and print it.

Practice

1. Why are strings considered special objects in Java compared to other objects?

easy

A. Because Java stores string literals in a special pool to save memory

B. Because strings can only store numbers

C. Because strings are primitive data types

D. Because strings cannot be compared using methods

Solution

1. Step 1: Understand string storage in Java

   Java stores string literals in a special memory area called the string pool to reuse them and save memory.

2. Step 2: Compare with other objects

   Other objects do not have this special pool and are stored separately each time they are created.

3. Final Answer:

   Because Java stores string literals in a special pool to save memory -> Option A

4. Quick Check:

   String pool = special memory reuse [OK]

Hint: Remember: string literals share memory in a pool [OK]

Common Mistakes:

• Thinking strings are primitive types
• Believing strings only store numbers
• Assuming strings cannot be compared

2. Which of the following is the correct way to create a string literal in Java?

easy

A. String s = new String('hello');

B. String s = "hello";

C. String s = 'hello';

D. String s = hello;

Solution

1. Step 1: Identify correct string literal syntax

   In Java, string literals must be enclosed in double quotes, like "hello".

2. Step 2: Check other options for errors

   String s = new String('hello'); uses single quotes which are for chars, String s = 'hello'; uses single quotes incorrectly, String s = hello; lacks quotes entirely.

3. Final Answer:

   String s = "hello"; -> Option B

4. Quick Check:

   Strings use double quotes "" [OK]

Hint: Use double quotes for strings, single for chars [OK]

Common Mistakes:

• Using single quotes for strings
• Omitting quotes around text
• Confusing char and string syntax

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

String a = "hello";
String b = "hello";
System.out.println(a == b);

medium

A. false

B. Runtime exception

C. Compilation error

D. true

Solution

1. Step 1: Understand string literal pooling

   Both variables a and b refer to the same string literal "hello" in the string pool, so they point to the same object.

2. Step 2: Evaluate '==' operator on strings

   The '==' operator compares references, and since both refer to the same object, it returns true.

3. Final Answer:

   true -> Option D

4. Quick Check:

   Same literal references = true [OK]

Hint: == true if both strings share the same pool object [OK]

Common Mistakes:

• Assuming '==' compares string content
• Expecting false because of separate variables
• Confusing '==' with equals() method

4. Identify the error in the following code snippet:

String s1 = new String("test");
String s2 = new String("test");
if (s1 == s2) {
    System.out.println("Equal");
} else {
    System.out.println("Not Equal");
}

medium

A. Using '==' to compare string contents instead of equals()

B. Missing semicolon after string declarations

C. Incorrect string initialization syntax

D. Cannot create strings with new keyword

Solution

1. Step 1: Understand '==' vs equals() for strings

   '==' compares references, not content. Since s1 and s2 are different objects, '==' returns false.

2. Step 2: Correct way to compare string content

   Use s1.equals(s2) to compare the actual text inside the strings.

3. Final Answer:

   Using '==' to compare string contents instead of equals() -> Option A

4. Quick Check:

   Use equals() for string content comparison [OK]

Hint: Use equals() to compare string text, not '==' [OK]

Common Mistakes:

• Using '==' to compare string values
• Thinking new String() creates same object
• Ignoring string pool behavior

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

String s1 = "Java";
String s2 = new String("Java");
String s3 = s2.intern();
System.out.println(s1 == s2);
System.out.println(s1 == s3);

Why does this happen?

hard

A. false false - all are different objects

B. true false - new String creates pooled string

C. false true - intern() returns the pooled string reference

D. true true - all strings are pooled automatically

Solution

1. Step 1: Analyze s1 and s2 references

   s1 is a string literal in the pool, s2 is a new String object outside the pool, so s1 == s2 is false.

2. Step 2: Understand intern() method effect

   s3 = s2.intern() returns the reference from the string pool for "Java", which is the same as s1, so s1 == s3 is true.

3. Final Answer:

   false true - intern() returns the pooled string reference -> Option C

4. Quick Check:

   intern() returns pool reference [OK]

Hint: intern() returns pooled string reference [OK]

Common Mistakes:

• Assuming new String() returns pooled string
• Thinking '==' compares string content
• Ignoring intern() method purpose