扫描程序以获取长整数，线程"main"中的异常； java.util.InputMismatchException [英] Scanner for long integer, Exception in thread "main" java.util.InputMismatchException

问题描述

我正在最后确定程序，但是，每当我输入整数(长整数)时，我都会遇到输入不匹配的情况:

I am at the last step to finalize my program, however whenever I enter my integer (long) I get a input mismatch:

Compiler message: "Exception in thread "main" java.util.InputMismatchException: For input               string: "4388576018402626"
at java.util.Scanner.nextInt(Scanner.java:2097)
at java.util.Scanner.nextInt(Scanner.java:2050)
at CreditCardValidation.main(CreditCardValidation.java:12)"

我的代码如下:

import java.util.Scanner ; //import Scanner

public class CreditCardValidation {

public static void main (String[] args){

    Scanner kbd = new Scanner(System.in) ;

    System.out.println("Please enter Creditcard number: " ) ;

    int credNumber = kbd.nextInt() ;

    boolean n = isValid( credNumber ) ;

if (credNumber == 0 )
    System.exit(0) ;

do {                

    System.out.println("Please enter Creditcard number: " ) ;
    credNumber = kbd.nextInt() ;
    }
    while ( credNumber < 0 ) ;

if (credNumber > 0 )
    System.out.print("This credit card number is " + n ) ;




}

/*
Return true is the number is a valid card number.
*/

public static boolean isValid(long number) {


    long p = getPrefix(number, 1);
    long p2 = getPrefix(number, 2);
    int n = getSize(number);


    if ((p == 4 || p == 5 || p == 6 || p2 == 37)&& (n < 13 || n > 16) && (((sumOfDoubleEvenPlace(number) + sumOfoddPlace(number))) % 10) == 0)
        return true ;
    else
        return false ;

}


/* The sum of every other digit, doubled, starting with the first digit. */

public static int sumOfDoubleEvenPlace(long number) {
    int sum = 0;
    int maxDigitLenth = 16;
    for (int i = 1; i <= maxDigitLenth; i++)
    {
        if (i % 2 == 0)
        {
            sum = sum + getDigit((int)(number % 10) * 2);
        }
        number /= 10;
    }
    return sum;
}

/*
Return the number if it is 0-9, otherwise return the sum of
the digits of the number.
*/

public static int getDigit(int number) {
    if (number < 10) {

        return number;
    }
    else {

        return (number / 10) + (number % 10);
    }

    }
/*
Return the sum of the odd-place digits.
*/

public static int sumOfoddPlace(long number) {
    int maxDigitLength = 16;
    int sum = 0;
    for (int i = 1; i <= maxDigitLength; i++)
    {

        if (i % 2 == 1)
        {
            sum = sum + (int)(number % 10);
        }
       number /= 10;
    }
    return sum;
}

/*
Return the number of digits in d
*/

public static int getSize(long d) {

    int size = 0 ;

    while( d > 0 ) {
        d = d / 10 ;
        size = size + 1 ;       
    }
    return size ;
}

/*
Return the first k number of digits from number. If the number of digits in number is 
less than k, return the number.
*/

public static long getPrefix(long n, int k) {

    int f = getSize(n)-k;

    long prefix = n/((long)(Math.pow(10, f)));

    return prefix;
    }

/*
Return true if the digit d is a prefix for number.
*/

public static boolean prefixMatched( long number, int d ) {

    if ( d == getPrefix(number, 4))

        return true ;
    else
        return false ;

    }

}

谢谢您的时间！

推荐答案

这是因为您输入的值超出了整数值的范围.在这种情况下，您需要使用很长时间.整数的最大值为 2147483647 .

That is because the value you're entering is beyond the range of integer values. You need to use long in this case. The max value of integer is 2147483647.

long credNumber = kbd.nextLong();
..
// in the do while loop also
credNumber = kbd.nextLong() ;

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