Question

我对编程非常陌生，因此，请帮助我解决Luhn的算法问题。一切可以正常工作，但签证为411111111111111。实际上是什么导致了错误？我知道复制粘贴被认为是错误的代码，但我试图至少创建有效的代码。你能帮助我吗改善此代码？感谢您的帮助。

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   #include <cs50.h>
   #include <stdio.h>

    int main ()
     { // 1

       long cardnumber;
       int digits = 0;
       int even, odd;

        //  Getting credit card number
       do 
        {
        cardnumber = get_long("Enter your card number:\n"); 
        }
        while (cardnumber < 0);

          //  calculates the total digits in card
  long digitcounter = cardnumber;
  while(digitcounter > 0)
  {
    digitcounter = digitcounter / 10;
    digits++;
  }

  if (digits != 13 &&  digits != 15 && digits != 16)
  {   // 2-2
    printf("INVALID");
  }    // 2-2
  else
    { // 2

    //  Breaks down the credit card number into individual integers
  int digit1 = (cardnumber / 1000000000000000);
  int digit2 = (cardnumber / 100000000000000) % 10;
  int digit3 = (cardnumber / 10000000000000) % 10;
  int digit4 = (cardnumber / 1000000000000) % 10;
  int digit5 = (cardnumber / 100000000000) % 10;
  int digit6 = (cardnumber / 10000000000) % 10;
  int digit7 = (cardnumber / 1000000000) % 10;
  int digit8 = (cardnumber / 100000000) % 10;
  int digit9 = (cardnumber / 10000000) % 10;
  int digit10 = (cardnumber / 1000000) % 10;
  int digit11 = (cardnumber / 100000) % 10;
  int digit12 = (cardnumber / 10000) % 10;
  int digit13 = (cardnumber / 1000) % 10;
  int digit14 = (cardnumber / 100) % 10;
  int digit15 = (cardnumber / 10) % 10;
  int digit16 = (cardnumber % 10);

  //  multiplies 2 to every odd number 
  int multi_odd15 = (digit15 * 2);
    while (multi_odd15 > 9)
    {
        int check1 = multi_odd15 % 10;
        int check2 = multi_odd15 / 10;
        multi_odd15 = check1 + check2;
    }
  int multi_odd13 = (digit13 * 2);
    while (multi_odd13 > 9)
    {
        int check1 = multi_odd13 % 10;
        int check2 = multi_odd13 / 10;
        multi_odd13 = check1 + check2;
    }
  int multi_odd11 = (digit11 * 2);
    while (multi_odd11 > 9)
    {
        int check1 = multi_odd11 % 10;
        int check2 = multi_odd11 / 10;
        multi_odd11 = check1 + check2;
    }
   int multi_odd9 = (digit9 * 2);
    while (multi_odd9 > 9)
    {
        int check1 = multi_odd9 % 10;
        int check2 = multi_odd9 / 10;
        multi_odd9 = check1 + check2;
    }
   int multi_odd7 = (digit7 * 2);
    while (multi_odd7 > 9)
    {
        int check1 = multi_odd7 % 10;
        int check2 = multi_odd7 / 10;
        multi_odd7 = check1 + check2;
    }
  int multi_odd5 = (digit5 * 2);
    while (multi_odd5 > 9)
    {
        int check1 = multi_odd5 % 10;
        int check2 = multi_odd5 / 10;
        multi_odd5 = check1 + check2;
    }
  int multi_odd3 = (digit3 * 2);  
    while (multi_odd3 > 9)
    {
        int check1 = multi_odd3 % 10;
        int check2 = multi_odd3 / 10;
        multi_odd3 = check1 + check2;
    }
  int multi_odd1 = (digit1 * 2);
    while (multi_odd1 > 9)
    {
        int check1 = multi_odd1 % 10;
        int check2 = multi_odd1 / 10;
        multi_odd1 = check1 + check2;
    }

    //sum of odd number
   odd = (multi_odd15 + multi_odd13 + multi_odd11 + multi_odd9 + multi_odd7 + multi_odd5 + multi_odd3 
       + multi_odd1);

    //  Sum of all the even numbers
   even = (digit2 + digit4 + digit6 + digit8 + digit10 + digit12 + digit14 + digit16);

    //  Code to check if the card number is possible
   if ((odd + even) % 10 == 0)
   { // 3    
    //  If the card number is 13 digits
    if (digits == 13)
    { // 4
        int checkvisa1 =  cardnumber / 1000000000000;
        if (checkvisa1 == 4)
        {
            printf("Visa\n");
        }
        else 
        {
            printf("INVALID\n");
        }
      } // 4  
    } //3

    //  If the card number is 15 digits
    else if (digits == 15)
    { // 5
        int checkamex = cardnumber / 10000000000000;
        if (checkamex == 34 && checkamex == 37)
        {
            printf("American express\n");
        }
        else 
        {
            printf("INVALID\n");
        }
    } // 5

    //  If the card number is 16 digits
    else
    { // 6
        int checkmaster = cardnumber / 100000000000000;
        int checkvisa2 = cardnumber / 1000000000000000;

        if (checkmaster <= 55 && checkmaster > 50)
        {
           printf("Mastercard\n");
        }
        else if (checkvisa2 == 4)
        {
           printf("Visa\n");
        }
        else
        {
           printf("INVALID\n");
        }
      } // 6

     } // 2

   } // 1

Answer 1

仅当 else if (digits == 15)为false时，才会执行此if ((odd + even) % 10 == 0)。那可能不是您想到的流量控制。更有可能应该与此if (digits == 13)“配对”。

当此if ((odd + even) % 10 == 0)为true时，下一个测试为if (digits == 13)，在4111111111111111示例中为false，因此不创建任何输出，并使用该块完成。 else if和下一个else块不会执行（因为luhn's test评估为true），所以程序只是...结束。

为什么VISA（4111111111111111）不返回任何内容？

1 个答案: