在卢恩的阿尔格号上写了个剧本。无法理解如何为修改后的脚本创建和调用函数

Question

我对编程很陌生。我目前正在进行非常严格的在线讲座，并使用Luhn的算法完成了一项任务。这只是一个脚本，直接贯穿，但对于我的未来，我想学习更有效的代码，因为项目变得更大。

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”的内容。

(根据机器人的说法，一切都已经提交，我的脚本已经完美地完成了任务，所以我不想半途而废。)

这是完成的脚本，只有主要的功能，直接运行，我花了大约12-15个小时，使它工作，从一开始就没有错误。所有东西都是用C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);
    }
}

我试着把它分成三种功能，主叫。

长龙input_number();

• bool luhn_check();

• void company_check();

我被第二个函数困住了，不确定第三个函数是否应该是空的。

“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);
    }
}

我已经完成了脚本的第二个版本和您的建议，除了输入的str之外，我将尝试在下一个版本中处理该方法，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，还有什么事情我可以做得更有效吗？

Answer 1

让我从关于您的函数input_number的几个注释开始

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
}