NumberFormatException in Java: Causes, Fixes, Examples & Interview Tips

What is This Error?

The NumberFormatException is thrown when you try to convert a String to a numeric type (int, long, double, etc.) but the string doesn't contain a valid number. It's a subclass of IllegalArgumentException.

Common Causes

String contains letters or special characters
Parsing a decimal string as integer (e.g., "12.5" with parseInt)
String has leading/trailing whitespace
Empty string or null passed to parse method
Number exceeds the type's range (overflow)

Quick Fix (TL;DR)

Quick Solution

// ❌ Problem
int num = Integer.parseInt("abc"); // NumberFormatException!

// ✅ Solution 1: try-catch
try {
    int num = Integer.parseInt(input.trim());
} catch (NumberFormatException e) {
    System.out.println("Invalid number: " + input);
}

// ✅ Solution 2: Validate first
if (input.matches("-?\\d+")) {
    int num = Integer.parseInt(input);
}

Common Scenarios & Solutions

Problem

Scanner scanner = new Scanner(System.in);
System.out.print("Enter age: ");
String input = scanner.nextLine();
int age = Integer.parseInt(input); // Crashes if user types "twenty"!

Solution

Scanner scanner = new Scanner(System.in);
int age = -1;

while (age < 0) {
    System.out.print("Enter age: ");
    String input = scanner.nextLine().trim();
    try {
        age = Integer.parseInt(input);
        if (age < 0 || age > 150) {
            System.out.println("Please enter a valid age (0-150)");
            age = -1;
        }
    } catch (NumberFormatException e) {
        System.out.println("Invalid input. Please enter a number.");
    }
}
System.out.println("Age: " + age);

Problem

String price = "19.99";
int p = Integer.parseInt(price); // NumberFormatException! "19.99" is not an int

Solution

String price = "19.99";

// ✅ Parse as double
double p = Double.parseDouble(price); // 19.99

// ✅ Or parse as int after truncating
int pInt = (int) Double.parseDouble(price); // 19

// ✅ Use BigDecimal for money
BigDecimal bd = new BigDecimal(price); // Exact representation

Problem

String input = " 42 "; // Has spaces!
int num = Integer.parseInt(input); // NumberFormatException!

Solution

String input = " 42 ";
int num = Integer.parseInt(input.trim()); // ✅ Always trim first!

Solution

// ✅ Reusable safe parse method
public static Optional<Integer> safeParseInt(String s) {
    if (s == null || s.trim().isEmpty()) return Optional.empty();
    try {
        return Optional.of(Integer.parseInt(s.trim()));
    } catch (NumberFormatException e) {
        return Optional.empty();
    }
}

// Usage
safeParseInt("42").ifPresent(n -> System.out.println("Parsed: " + n));
safeParseInt("abc").ifPresentOrElse(
    n -> System.out.println(n),
    () -> System.out.println("Invalid number")
);

Best Practices to Avoid This Error

Always trim() before parsing - Remove leading/trailing whitespace
Validate with regex before parsing - Use matches("-?\d+") for integers
Use try-catch for user input - Users can type anything
Use the right parse method - parseInt for int, parseDouble for decimals
Create safe parse helper methods - Reusable validation logic
Use BigDecimal for money - Avoids floating-point precision issues
Check for null/empty before parsing - Prevent NullPointerException too

Related Errors

Frequently Asked Questions

What causes NumberFormatException?

It's thrown when Integer.parseInt(), Double.parseDouble(), or similar methods receive a string that doesn't represent a valid number "” like "abc", "12.5" (for parseInt), or strings with spaces.

How do I parse "12.5" as an integer?

First parse as double, then cast: (int) Double.parseDouble("12.5") gives 12. Or use Math.round() to round: (int) Math.round(Double.parseDouble("12.5")) gives 13.

How do I validate a string is a number before parsing?

Use regex: input.matches("-?\d+") for integers, input.matches("-?\d+(\.\d+)?") for decimals. Or use try-catch which is simpler for most cases.

What's the difference between parseInt and valueOf?

Integer.parseInt() returns a primitive int. Integer.valueOf() returns an Integer object (autoboxed). Both throw NumberFormatException for invalid input.

How do I handle NumberFormatException from user input?

Use a try-catch loop that keeps asking until valid input is provided. Always trim() the input first. Provide clear error messages telling the user what format is expected.

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NumberFormatException in Java: Causes, Fixes, Examples & Interview Tips

What is This Error?

Common Causes

Quick Fix (TL;DR)

Quick Solution

Common Scenarios & Solutions

Problem

Solution

Problem

Solution

Problem

Solution

Solution

Best Practices to Avoid This Error

Related Errors

Frequently Asked Questions

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