c - 在 Luhn's Alg 上制作了一个脚本。无法清楚地理解如何为修改后的脚本创建和调用函数

Question

我是编程新手。我目前正在非常严格地接受在线讲座，并使用卢恩算法完成了一项任务。这只是一个直接运行的脚本，但对于 future 的我来说，我想随着项目变得越来越大，学习更有效地编码。

这就是我的问题所在。我似乎无法理解如何正确定义或调用函数，也无法将我的脚本修改为更“高效”的东西。

(根据机器人的说法，一切都已经提交，我的脚本完美地完成了任务，所以我不想在这里半途而废我的工作只是为了清楚起见。)

这是完整的脚本，仅包含直接运行的 main 函数，我花了大约 12-15 个小时才让它从一开始就没有错误地运行。一切都是用 C 编写的

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

int main(void)
{
    // this grabs the number and verifies the correct amount of digits
    int count;
    long number = 0;
    do
    {
        number = get_long("Number: ");
        count = (number == 0) ? 1 : (log10(number) + 1);
        if (count < 13 || count == 14 || count > 16)
        {
            printf("INVALID\n"); // This is to satisfy the uni. bot command check. Need a EOF for numbers with the wrong amount of digits.
            return (0);
        }
    }
    while (count < 13 || count == 14 || count > 16);
    //printf("Digits: %i\n", count);       // test print for debugging

    //This finds the first two digits of input number

    long int two = number;
    while (two >= 100)
    {
        two = two / 10;
    }
    //printf("First two numbers: %li\n", two);      // test print for debugging

    // verifies card using mod10 (Luhn's)
    long sum = 0;
    long bigdigit = 0;
    //printf("\nLUHN Number: %li\n\n", number);     // test print for debugging
    if (count == 13 || count == 15)
    {
        count += 1;
    }
    for (int i = count; i > 0; i--)
    {
        if (i % 2 == 0)
        {
            sum += (number % 10);
        }
        else
        {
            bigdigit = (2 * (number % 10));
            sum += (bigdigit / 10 + bigdigit % 10);
        }
        number = (number / 10);
        //printf("\nI : %i\n", i);      // test print for debugging
        //printf("Sum: %li\n", sum);        // test print for debugging
        //printf("Number: %li\n", number);      // test print for debugging
    }
    if (sum % 10 == 0)
    {
        printf("VALID\n");
    }
    else
    {
        printf("INVALID\n");
        return (0);
    }

    // checks what type of card

    if (two == 34 || two == 37)
    {
        printf("AMEX\n");
        return (0);
    }
    else if (two >= 51 && two <= 55)
    {
        printf("MASTERCARD\n");
        return (0);
    }
    else if (two >= 40 && two <= 49)
    {
        printf("VISA\n");
        return (0);
    }
    else
    {
        printf("INVALID\n");
        return (0);
    }
}

我试图将它分成 3 个 main 函数来调用。

• long input_number();
• bool luhn_check();
• 无效company_check();

我被第二个函数困住了，不确定第三个函数是否应该为空。

<小时/>

“修订”脚本 v2

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

long input_number(long CCN);
int counter(long CCN, int count);
bool luhn_check(long CCN, int count);
long firsttwo(long CCN, long two);
void card_type(long two);


int main()
{
    long CCN = 0;
    int count = 0;
    long two = 0;
    CCN = input_number(CCN);
    count = counter(CCN, count);

    //printf("CCN: %li\n", CCN); //debugging purposes
    //printf("Digits: %i\n", count); //debugging purposes
    luhn_check(CCN, count);
    two = firsttwo(CCN, two);
    //printf("First Two: %li\n", two); //debugging purposes
    card_type(two);
}


    // this grabs the number and verifies the correct amount of digits
long input_number(long CCN)
{
    int count = 0;
    do
    {
        CCN = get_long("Number: ");
        count = (CCN == 0) ? 1 : (log10(CCN) + 1);
        if (count < 13 || count == 14 || count > 16)
        {
            //printf("INVALID\n"); // This is to satisfy the uni. bot command check. Need a EOF
            //return (0);
        }
    }
    while (count < 13 || count == 14 || count > 16);   
    return (CCN);
}

int counter(long CCN, int count)
{
 do
    {
        count = (CCN == 0) ? 1 : (log10(CCN) + 1);
    }
    while (count < 13 || count == 14 || count > 16);   
    return (count);   
}

    // verifies card using mod10 (Luhn's)
bool luhn_check(long CCN, int count)
{
    long sum = 0;
    long bigdigit = 0;
    //printf("\nLUHN Number: %ld\n\n", CCN);     // test print for debugging
    if (count == 13 || count == 15)
    {
        count += 1;
    }
    for (int i = count; i > 0; i--)
    {
        if (i % 2 == 0)
        {
            sum += (CCN % 10);
        }
        else
        {
            bigdigit = (2 * (CCN % 10));
            sum += (bigdigit / 10 + bigdigit % 10);
        }
        CCN = (CCN / 10);
        //printf("\nI : %i\n", i);      // test print for debugging
        //printf("Sum: %li\n", sum);        // test print for debugging
        //printf("Number: %li\n", CCN);      // test print for debugging
    }
    if (sum % 10 == 0)
    {
        printf("VALID\n");
        return (true);
    }
    else
    {
        printf("INVALID\n");
        return (false);
    }
}

            // grabs the first two numbers
long firsttwo(long CCN, long two)
{
    two = CCN;
    //printf("TWO CCN: %li\n", two); // debugging purposes
    while (two >= 100)
    {
        two = two / 10;
    }
    return (two);
}

            // finds card type and ends
void card_type(long two)
{
    if (two == 34 || two == 37)
    {
        printf("AMEX\n");
        //return (0);
    }
    else if (two >= 51 && two <= 55)
    {
        printf("MASTERCARD\n");
        //return (0);
    }
    else if (two >= 40 && two <= 49)
    {
        printf("VISA\n");
        //return (0);
    }
    else
    {
        printf("INVALID\n");
        //return (0);
    }
}

我已经根据您的建议完成了脚本的第二个版本，除了输入字符串，我将尝试在下一个版本中解决该方法，因为我还没有处理该类型。

除了使用字符串而不是b运行程序之外，还有什么我可以更有效地完成的吗？

Answer 1

让我先对您的函数 input_number 做一些说明:

long input_number()
{
    // this grabs the number and verifies the correct amount of digits
    int count;
    [ .... ]

    while (count < 13 || count == 14 || count > 16);
    // Once this while loop starts, it will never terminate!
    // if count starts at 10 (which is less than 13), the body says "do nothing".
    // so count will continue to be unchanged, and the while-loop will continue to run.

    return (0);  // Code after a return statement will never be reached
                 // So your printf-statement below will NEVER happen.
                 // It also seems unlikely you want to return Zero.
                 // You probably want to return count, or some other value.

    printf("Digits: %i\n", count);       // test print for debugging
}