String creation in Java - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Java code?

Consider the following Java code snippet:

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

What will be printed?

Java

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

A false

B true

C Compilation error

D Runtime exception

💡 Hint

Remember that == compares references, not content.

The == operator checks if both variables point to the same object. Here, s1 points to a string literal in the string pool, while s2 points to a new String object. So, the result is false.

❓ Predict Output

intermediate

What is the output of this Java code using StringBuilder?

Look at this code:

StringBuilder sb = new StringBuilder("abc");
sb.append("def");
System.out.println(sb.toString());

What will be printed?

Java

StringBuilder sb = new StringBuilder("abc");
sb.append("def");
System.out.println(sb.toString());

A abc def

B abcdef

C abc

D Compilation error

💡 Hint

Think about what append does to the StringBuilder.

The append method adds the given string to the end of the current content. So, sb becomes "abcdef".

🔧 Debug

advanced

What error does this code raise?

Examine this Java code snippet:

String s = null;
int length = s.length();
System.out.println(length);

What error will this code cause when run?

Java

String s = null;
int length = s.length();
System.out.println(length);

A NullPointerException

B Compilation error

C ArrayIndexOutOfBoundsException

D No error, prints 0

💡 Hint

Think about calling a method on a null reference.

Calling length() on a null String causes a NullPointerException at runtime.

❓ Predict Output

advanced

What is the output of this code using String intern()?

Consider this Java code:

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

What will be printed?

Java

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

A Compilation error

B false

C Runtime exception

D true

💡 Hint

Remember what intern() does to a String.

The intern() method returns a reference to the string pool object. Since s3 is the literal "test" in the pool, s2 == s3 is true.

🧠 Conceptual

expert

How many String objects are created here?

Analyze this Java code snippet:

String s = "a" + "b" + new String("c");

How many distinct String objects are created at runtime?

A 4

B 3

C 2

D 1

💡 Hint

Consider compile-time constants and new String() usage.

The literals "a" and "b" are combined at compile time into "ab" (one object in the pool). The new String("c") creates a new object at runtime. The concatenation creates a new String object at runtime. So total runtime objects created are 2: the new String("c") and the concatenation result.

Practice

1. Which of the following is the correct way to create a string variable in Java?

easy

A. String greeting = "Hello";

B. string greeting = 'Hello';

C. String greeting = Hello;

D. String greeting = "Hello"

Solution

1. Step 1: Check the data type and quotes

   Java strings use the class String with a capital S and text must be in double quotes.

2. Step 2: Validate syntax correctness

   String greeting = "Hello"; uses correct capitalization, double quotes, and ends with a semicolon.

3. Final Answer:

   String greeting = "Hello"; -> Option A

4. Quick Check:

   String creation uses String and double quotes [OK]

Hint: Use capital S in String and double quotes for text [OK]

Common Mistakes:

• Using single quotes instead of double quotes
• Using lowercase 'string' instead of 'String'
• Missing semicolon at the end

2. Which of the following lines will cause a syntax error in Java?

easy

A. String name = "Alice";

B. String name = 'Alice';

C. String name = "Bob";

D. String name = "Charlie";

Solution

1. Step 1: Identify correct string literal syntax

   Java strings must be enclosed in double quotes, not single quotes.

2. Step 2: Check each option

   String name = 'Alice'; uses single quotes which is invalid for strings in Java, causing a syntax error.

3. Final Answer:

   String name = 'Alice'; -> Option B

4. Quick Check:

   Strings need double quotes, single quotes cause syntax error [OK]

Hint: Remember: Java strings use double quotes only [OK]

Common Mistakes:

• Using single quotes for strings
• Confusing char literals with strings
• Missing semicolon

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

String first = "Good";
String second = "Morning";
System.out.println(first + " " + second);

medium

A. GoodMorning

B. Good+Morning

C. Good Morning

D. gninroM dooG

Solution

1. Step 1: Understand string concatenation

   The + operator joins strings. Here, first + " " + second adds a space between words.

2. Step 2: Combine the strings

   "Good" + " " + "Morning" results in "Good Morning" with a space.

3. Final Answer:

   Good Morning -> Option C

4. Quick Check:

   + joins strings including spaces [OK]

Hint: Use + with " " to add spaces between words [OK]

Common Mistakes:

• Forgetting to add space between words
• Using + without quotes for space
• Expecting + to print plus sign

4. Find the error in the following code snippet:

String text = "Hello"
System.out.println(text);

medium

A. System.out.println syntax is wrong

B. Incorrect string quotes

C. Variable name is invalid

D. Missing semicolon after string assignment

Solution

1. Step 1: Check string assignment line

   The line ends without a semicolon, which is required in Java statements.

2. Step 2: Verify other lines

   Quotes are correct, variable name is valid, and print statement syntax is correct.

3. Final Answer:

   Missing semicolon after string assignment -> Option D

4. Quick Check:

   Every statement must end with a semicolon [OK]

Hint: Always end Java statements with a semicolon [OK]

Common Mistakes:

• Omitting semicolon at end of statement
• Confusing quotes with apostrophes
• Misnaming variables

5. You want to create a string variable that holds the full name by combining first and last names with a space. Which code correctly does this?

hard

A. String fullName = firstName + " " + lastName;

B. String fullName = firstName + lastName;

C. String fullName = firstName.concat(lastName);

D. String fullName = firstName - lastName;

Solution

1. Step 1: Understand string concatenation with space

   To add a space between first and last names, include " " between them.

2. Step 2: Check each option

   String fullName = firstName + " " + lastName; correctly uses + operator with a space string. String fullName = firstName + lastName; misses space, C concatenates without space, D uses invalid operator.

3. Final Answer:

   String fullName = firstName + " " + lastName; -> Option A

4. Quick Check:

   Use + " " + to add space between strings [OK]

Hint: Add " " between strings to insert space [OK]

Common Mistakes:

• Forgetting space between names
• Using invalid operators like -
• Using concat() without space