Parsing numeric arguments in Java - Mini Project: Build & Apply

Parsing numeric arguments

📖 Scenario: You are creating a simple Java program that reads two numbers given as text (strings) and converts them into numbers to add them together.

🎯 Goal: Build a Java program that parses two string arguments into integers, adds them, and prints the result.

📋 What You'll Learn

Create two string variables with exact values "15" and "27"

Create an integer variable to hold the parsed value of the first string

Create an integer variable to hold the parsed value of the second string

Add the two integers and store the result in a variable

Print the result

💡 Why This Matters

🌍 Real World

Parsing numeric arguments is common when reading user input, command-line arguments, or data from files where numbers are stored as text.

💼 Career

Many programming jobs require converting text input into numbers to perform calculations, validations, or data processing.

1

Create string variables for the numbers

Create two string variables called num1 and num2 with the exact values "15" and "27" respectively.

Java

// Create two string variables num1 and num2 with values "15" and "27"
// Your code here

Hint

Use String type and assign the exact text values.

2

Parse the strings into integers

Add two integer variables called number1 and number2 that parse the strings num1 and num2 using Integer.parseInt().

Java

public class Main {
    public static void main(String[] args) {
        String num1 = "15";
        String num2 = "27";
        // Parse num1 and num2 into integers number1 and number2
        // Your code here
    }
}

Hint

Use Integer.parseInt() to convert strings to integers.

3

Add the two integers

Create an integer variable called sum that adds number1 and number2.

Java

public class Main {
    public static void main(String[] args) {
        String num1 = "15";
        String num2 = "27";
        int number1 = Integer.parseInt(num1);
        int number2 = Integer.parseInt(num2);
        // Add number1 and number2 into sum
        // Your code here
    }
}

Hint

Use the + operator to add two integers.

4

Print the sum

Use System.out.println() to print the value of sum.

Java

public class Main {
    public static void main(String[] args) {
        String num1 = "15";
        String num2 = "27";
        int number1 = Integer.parseInt(num1);
        int number2 = Integer.parseInt(num2);
        int sum = number1 + number2;
        // Print the sum
        // Your code here
    }
}

Hint

Use System.out.println(sum); to display the result.

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