Parsing numeric arguments in Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the purpose of parsing numeric arguments in Java?

Parsing numeric arguments means converting strings (like command line inputs) into numbers (like int or double) so the program can use them in calculations.

beginner

Which Java method converts a String to an int?

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

intermediate

How do you safely handle invalid numeric input when parsing in Java?

Use a try-catch block to catch NumberFormatException. This prevents the program from crashing if the input is not a valid number.

beginner

What is the difference between Integer.parseInt() and Double.parseDouble()?

Integer.parseInt() converts a string to an integer (whole number), while Double.parseDouble() converts a string to a double (decimal number).

intermediate

Why is it important to validate numeric arguments after parsing?

Because even if parsing succeeds, the number might be out of expected range or invalid for the program logic. Validation ensures the program behaves correctly.

Which Java method converts a string "123" to the integer 123?

A Integer.toString(123)

B Double.parseDouble("123")

C String.valueOf(123)

D Integer.parseInt("123")

Integer.parseInt converts a string to an int. Double.parseDouble converts to a double. String.valueOf and Integer.toString convert numbers to strings.

What exception should you catch when parsing a number from a string in Java?

A NumberFormatException

B IOException

C NullPointerException

D ArithmeticException

NumberFormatException is thrown when a string cannot be converted to a number.

Which method would you use to parse a decimal number from a string?

A Double.parseDouble()

B String.parseDouble()

C Integer.parseInt()

D Float.toString()

Double.parseDouble converts a string to a double, which can hold decimal numbers.

What happens if you try to parse "abc" as an integer using Integer.parseInt("abc")?

A Returns 0

B Throws NumberFormatException

C Returns null

D Compiles error

Parsing a non-numeric string throws NumberFormatException at runtime.

Why should you validate numeric arguments after parsing?

A To improve performance

B To convert it back to string

C To check if the number fits program rules

D To avoid syntax errors

Validation ensures the number is within expected limits or conditions for correct program behavior.

Explain how to convert a command line argument to an integer in Java and handle possible errors.

Think about converting strings to numbers and what happens if the input is not a number.

Describe the difference between parsing integers and doubles from strings in Java.

Consider what kind of numbers each method handles.

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