NumberFormatException in Java — How to Fix (2026) | Tutorials Logic

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.

Error Message:
Exception in thread "main" java.lang.NumberFormatException: For input string: "abc"
java.lang.NumberFormatException: For input string: "12.5" (when parsing int)

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

Scenario 1: User Input Validation

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);

Scenario 2: Parsing Decimal as Integer

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

Scenario 3: Whitespace in String

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!

Scenario 4: Safe Parsing Helper Method

Solution

// ✅ Reusable safe parse method
public static Optional 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")
);