Question

我什至尝试使用 long int 代替 int ，但是编译器花费太多时间执行并且仍然不提供任何输出。

我正在使用CS50的IDE。

这是导致问题的代码：

int sum_alt = 0;
int sum_rest = 0;
int total = 0;

for(int i = 1; i < count; i=+2)
{
    int temp = number[i] * 2;
    sum_alt = sum_alt + temp;
}

for(int i = 0; i <= count; i=+2)
{
    int temp = number[i];
    sum_rest = sum_rest + temp;
}

total = sum_rest + sum_alt;

如何解决此问题？

这是完整的代码：

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

int main(void)
{
long long Cardnumber;

//Typos correction + User input
do
{
    Cardnumber = get_long_long("Number: ");
}
while(Cardnumber < 0);

int count = 0;
long long counter = Cardnumber;

//To count the number of digits
while(counter != 0)
{
    counter = counter / 10;
    count++;
}

//Declaring an array for cardnumber
printf("%i", count);
int number[count];

if((count != 13) && (count != 15) && (count != 16))
{
    printf("INVALID\n");
}

for(int access = 0; access < count; access++)
{
    number[access] = Cardnumber % 10;
    Cardnumber = Cardnumber / 10;
}

int originalnumber[count];

for(int i = 0; i < count; i++)
{
    originalnumber[i] = number[i];
}

long long int sum_alt = 0;
long long int sum_rest = 0;
long long int total = 0;

for(int i = 1; i < count; i=+2)
{
    int temp = number[i] * 2;
    sum_alt = sum_alt + temp;
}

for(int i = 0; i < count; i=+2)
{
    int temp = number[i];
    sum_rest = sum_rest + temp;
}

total = sum_rest + sum_alt;

if(total % 10 == 0)
{
    if(((count == 13) || (count == 16)))
    {
        if((originalnumber[12] == 4) || originalnumber[15] == 4)
        {
            printf("VISA\n");
        }
        else if(originalnumber[15]== 5)
        {
            if(originalnumber[14] == 1 || originalnumber[14]== 2 || originalnumber[14] == 3 || originalnumber[14] == 4 || originalnumber[14] == 5)
            {
                printf("MASTERCARD\n");
            }
        }

        else
        {
            printf("INVALID\n");
        }
    }
    else if (count == 15)
    {
        if(originalnumber[15] == 3)
        {
            if(originalnumber[15] == 4 || originalnumber[15] == 7)
            {
                printf("AMERICAN EXPRESS\n");
            }
            else
            {
                printf("INVALID\n");
            }
        }
        else
        {
            printf("INVALID\n");
        }
    }
}

else
{
    printf("INVALID\n");
}   

}

PS。我不知道CS50的IDE花这么多时间进行计算的原因。

Answer 1

这是因为2147483644 + 4 = 2147483648，它大于2147483647（有符号int32限制）。 64位整数类型为 long long int ，即AFAIK。

应该是：

long long int sum_alt = 0;
long long int sum_rest = 0;
long long int total = 0;

for(int i = 1; i < count; i=+2)
{
    int temp = number[i] * 2;
    sum_alt = sum_alt + temp;
}

for(int i = 0; i <= count; i=+2)
{
    int temp = number[i];
    sum_rest = sum_rest + temp;
}

total = sum_rest + sum_alt;

P.S。另外，该代码未定义变量“ count”，因此可能存在问题。

Answer 2

以下代码从用户那里获得信用卡号，检查其长度和校验和，然后确定发行者。

可以使用从https://ccardgenerator.com/获得的信用卡号对程序进行测试。

这段代码的属性是

它没有输入可能确实溢出的整数
它处理字符序列
正确实施了Luhn算法（必须对number[i]*2的数字求和）
我改进了测试以识别信用卡发卡行（基于前4位数字）；原始代码中有很多错误

#include <stdio.h>
#include <ctype.h>

int main()
{
    char buf[32];
    printf("enter credit card number: ");
    fgets(buf, sizeof(buf), stdin);

    int len = 0;
    while(isdigit(buf[len]))
        len++;
    if(len < 13 || len > 16) {
        printf("INVALID\n");
        return 1;
    }

    // verify credit card number check sum
    int checkSum = 0;
    for(int i = 0; i < len; i++) {
        int d = buf[len-i-1] - '0'; // get digit from right most to left most 
        if((i&1) == 0)
            checkSum += d;
        else
            checkSum += (d*2)%10 + (d*2)/10;
    }
    if(checkSum%10 != 0) {
        printf("INVALID\n");
        return 1;
    }

    // determine issuer (see https://www.regular-expressions.info/creditcard.html)
    int start = (buf[0]-'0')*1000 + (buf[1]-'0')*100 + (buf[2]-'0')*10 + buf[3]-'0';
    if((len == 13 || len == 16) && start/1000 == 4) {
        printf("VISA\n");
    } else if(len == 16 && ((start >= 5100 && start < 5600) || (start >= 2221 && start <= 2720))) {
        printf("MASTERCARD\n");
    } else if(len == 15 && ((start >= 3400 && start < 3500) || (start >= 3700 && start < 3800))) {
        printf("AMERICAN EXPRESS\n");
    } else if(len == 14 && ((start >= 3000 && start < 3060) || start/100 == 36 || start/100 == 38)) {
        printf("DINERS CLUB\n");
    } else if(len == 16 && (start == 6011 || start/100 == 65)) {
        printf("DISCOVER\n");
    } else if((len == 15 && (start == 2131 || start == 1800)) || (len == 16 && start/100 == 35)) {
        printf("JCB\n");
    } else {
        printf("UNKNOWN\n");
    }

    return 0;
}

如何解决运行时错误问题，例如“有符号整数溢出：2147483644 + 4无法以'int'类型表示”？

2 个答案: