Parsing numeric arguments in Java - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of parsing a valid integer string?

Consider the following Java code snippet that parses a string to an integer. What will be printed?

Java

public class Main {
    public static void main(String[] args) {
        String input = "123";
        int number = Integer.parseInt(input);
        System.out.println(number);
    }
}

A 123

B NumberFormatException

C 0

D null

💡 Hint

Think about what Integer.parseInt does with a valid numeric string.

The method Integer.parseInt converts the string "123" to the integer 123, which is then printed.

❓ Predict Output

intermediate

What happens when parsing a non-numeric string?

What will be the output or result of running this Java code?

Java

public class Main {
    public static void main(String[] args) {
        String input = "abc";
        int number = Integer.parseInt(input);
        System.out.println(number);
    }
}

A 0

B abc

C null

D NumberFormatException

💡 Hint

What does Integer.parseInt do if the string is not a valid number?

Parsing a non-numeric string with Integer.parseInt throws a NumberFormatException.

❓ Predict Output

advanced

Parsing a floating-point string with Integer.parseInt

What will happen when this code runs?

Java

public class Main {
    public static void main(String[] args) {
        String input = "12.34";
        int number = Integer.parseInt(input);
        System.out.println(number);
    }
}

A 12.34

B 12

C NumberFormatException

D 0

💡 Hint

Integer.parseInt expects a string representing an integer only.

The string "12.34" is not a valid integer format, so Integer.parseInt throws NumberFormatException.

❓ Predict Output

advanced

Parsing a floating-point string with Double.parseDouble

What will be the output of this Java code?

Java

public class Main {
    public static void main(String[] args) {
        String input = "12.34";
        double number = Double.parseDouble(input);
        System.out.println(number);
    }
}

A NumberFormatException

B 12.34

C 12

D 0.0

💡 Hint

Double.parseDouble can parse decimal numbers from strings.

Double.parseDouble converts the string "12.34" to the double value 12.34, which is printed.

🧠 Conceptual

expert

What is the value of 'result' after parsing with radix 16?

Given the code below, what is the value of the variable 'result' after execution?

Java

public class Main {
    public static void main(String[] args) {
        String hex = "1A";
        int result = Integer.parseInt(hex, 16);
        System.out.println(result);
    }
}

A 26

B 1A

C 16

D NumberFormatException

💡 Hint

Hexadecimal 1A equals decimal 26.

Parsing "1A" as a base-16 number converts it to decimal 26.

Practice

1.

Which Java method is used to convert a String containing a whole number into an int?

easy

A. Integer.parseInt(String s)

B. Double.parseDouble(String s)

C. String.toInt()

D. Integer.toString(int i)

Solution

1. Step 1: Identify method for integer parsing

   The method Integer.parseInt(String s) converts a string to an int.

2. Step 2: Compare with other options

   Double.parseDouble parses decimals, String.toInt() does not exist, and Integer.toString converts int to string.

3. Final Answer:

   Integer.parseInt(String s) -> Option A

4. Quick Check:

   Whole number parsing = Integer.parseInt [OK]

Hint: Use Integer.parseInt for whole numbers [OK]

Common Mistakes:

• Using Double.parseDouble for integers
• Trying to call non-existent String.toInt()
• Confusing toString() with parsing

2.

Which of the following is the correct syntax to parse a decimal number from a String named input in Java?

double value = ?;

easy

A. Double.parseInt(input);

B. Integer.parseDouble(input);

C. Integer.parseInt(input);

D. Double.parseDouble(input);

Solution

1. Step 1: Identify method for parsing decimals

   Use Double.parseDouble(String s) to convert string to double.

2. Step 2: Check syntax correctness

   Only Double.parseDouble(input) matches correct method and syntax.

3. Final Answer:

   Double.parseDouble(input); -> Option D

4. Quick Check:

   Decimal parsing = Double.parseDouble [OK]

Hint: Use Double.parseDouble for decimals [OK]

Common Mistakes:

• Using Integer.parseInt for decimals
• Calling non-existent Double.parseInt
• Mixing Integer and Double parse methods

3.

What is the output of the following Java code?

String s = "123";
int num = Integer.parseInt(s);
System.out.println(num + 10);

medium

A. 12310

B. 133

C. Error: cannot parse string

D. 10

Solution

1. Step 1: Parse string to integer

   Integer.parseInt("123") converts string "123" to integer 123.

2. Step 2: Add 10 to parsed integer

   Adding 10 to 123 results in 133.

3. Final Answer:

   133 -> Option B

4. Quick Check:

   123 + 10 = 133 [OK]

Hint: Parsed number + 10 means numeric addition [OK]

Common Mistakes:

• Thinking output is string concatenation '12310'
• Assuming parseInt throws error on valid input
• Confusing print output with input string

4.

Identify the error in this Java code snippet that tries to parse a number:

String input = "12a3";
int value = Integer.parseInt(input);
System.out.println(value);

medium

A. Syntax error: missing semicolon

B. No error, prints 123

C. Throws NumberFormatException at runtime

D. Compiles but prints 0

Solution

1. Step 1: Analyze input string content

   The string "12a3" contains a non-digit character 'a', invalid for integer parsing.

2. Step 2: Understand Integer.parseInt behavior

   Parsing invalid numeric strings throws NumberFormatException at runtime.

3. Final Answer:

   Throws NumberFormatException at runtime -> Option C

4. Quick Check:

   Invalid numeric string causes NumberFormatException [OK]

Hint: Invalid chars cause NumberFormatException [OK]

Common Mistakes:

• Assuming parseInt ignores invalid chars
• Expecting compile-time error instead of runtime
• Thinking output will be 0 on error

5.

You want to safely parse a String input to an int in Java, returning -1 if the input is not a valid number. Which code snippet correctly implements this?

hard

A. int parseSafe(String s) {
try {
return Integer.parseInt(s);
} catch (NumberFormatException e) {
return -1;
}
}

B. int parseSafe(String s) {
return Integer.parseInt(s);
}

C. int parseSafe(String s) {
if (s.matches("\\d+")) return Integer.parseInt(s);
else return 0;
}

D. int parseSafe(String s) {
return Integer.valueOf(s);
}

Solution

1. Step 1: Understand safe parsing with exception handling

   Using try-catch around Integer.parseInt catches invalid input and returns fallback.

2. Step 2: Check each option's behavior

   The snippet with try-catch returns parsed int or -1 on error. Snippets without proper handling throw exceptions, return 0 instead of -1, or use valueOf which returns an Integer object not int.

3. Final Answer:

   The try-catch code returning -1 on NumberFormatException -> Option A

4. Quick Check:

   Try-catch with -1 fallback handles invalid input [OK]

Hint: Use try-catch to handle invalid input safely [OK]

Common Mistakes:

• Not handling NumberFormatException
• Returning wrong fallback value
• Confusing Integer.valueOf with parseInt