"如何检测一个数字是否大于 Long.MAX 值"

Question

My application will get number as string from end user. If the number is not numeric, i have to throw error message by saying that to provide number. This i can fix by using NumberFormatException. Another scenario is, user entered greater than Long.MAX value. How i can check this case and give error message to the user to enter smaller number than Long.MAX value? I should not use any third party or open source lib to fix this issue. Even if they are providing solution, How they are resolving it?

"

Answer 1

Use BigInteger to parse user input and compare the result with Long.MAX_VALUE

String userInput = ...;
BigInteger bigInt = new BigInteger(userInput);
if(bigInt.compareTo(BigInteger.valueOf(Long.MAX_VALUE)) > 0) {
    throw new Exception(userInput + ": value is too large");
}

Answer 2

If the entered number is greater than Long.MAX value, then what will you do next. It will cause an error as you don't know where to store it.

Better way is to check at the time of user input is in range or not. If it is greater than Long.MAX , store it in BigInteger

Answer 3

Use BigInteger and the longValueExact() method, and catch exceptions:

public static void main(String[] args) {
    test("123");
    test("9223372036854775807");  // Long.MAX_VALUE
    test("-9223372036854775808"); // Long.MIN_VALUE
    test("9223372036854775808");  // Long.MAX_VALUE + 1
    test("-9223372036854775809"); // Long.MIN_VALUE - 1
    test("abc");
}
private static void test(String input) {
    long longVal;
    try {
        longVal = new BigInteger(input).longValueExact();
    } catch (NumberFormatException e) {
        System.out.println("Value is not a valid integer number: " + input);
        return;
    } catch (ArithmeticException e) {
        System.out.println("Value exceeds range of long: " + input);
        return;
    }
    System.out.println("Got valid long value: " + longVal);
}

OUTPUT

Got valid long value: 123
Got valid long value: 9223372036854775807
Got valid long value: -9223372036854775808
Value exceeds range of long: 9223372036854775808
Value exceeds range of long: -9223372036854775809
Value is not a valid integer number: abc

Answer 4

您可以使用 Long.MAX_VALUE 访问最大值并检查用户在 if 条件中输入的值。

Answer 5

Here is another solution without using an extra class other than Java core

public static void main(String[] args) {
    System.out.println(isLargerThanLONGMAXVALUE("9223372036854775807"));      // false
    System.out.println(isLargerThanLONGMAXVALUE("9223372036854775806"));      // false
    System.out.println(isLargerThanLONGMAXVALUE("9223372036854775808"));      // true
    System.out.println(isLargerThanLONGMAXVALUE("645459223372036854775807")); // true
    System.out.println(isLargerThanLONGMAXVALUE("922"));                      // false
}

public static boolean isLargerThanLONGMAXVALUE (String number) {
    String longMax = String.valueOf(Long.MAX_VALUE);
    if (number.length() > longMax.length()) return true;
    if (number.length() < longMax.length()) return false;
    long a, b = 0;
    for (int i = 1 ; i < number.length() ; i++){
        a = Long.parseLong(number.substring(0, i));
        b = Long.parseLong(longMax.substring(0, i));
        if (a > b) return true;
    }
    if (Integer.parseInt(number.substring(number.length()-1, number.length())) > 
        Integer.parseInt(longMax.substring(number.length()-1, number.length())))
        return true;
    return false;
}

Answer 6

Treating the string as a BigInteger and doing the comparison is the best way. But here's another just to show that there's usually more than one way to accomplish something:

public boolean isInRange(String number) {
    String maxValue = Long.toString(Long.MAX_VALUE);
    number = number.replaceFirst("^0+", "");  // remove leading zeroes
    return number.length() < maxValue.length() ||
           (number.length() == maxValue.length() &&
            number.compareTo(maxValue) <= 0);
}

This assumes that number is composed entirely of digits (no negative sign).

Answer 7

try{
  val n = input.toLong()
}catch(e: Exception){
  // invalid Long
}