c - 基于结构的程序、链接器和多个定义？

Question

这是给我的文件，我被告知不要更改这两个文件中的任何一个。函数的说明在头文件中，main.c 是测试代码的 shell:

日期.h:

#ifndef _DateH_
#define _DateH_

typedef struct {
int month;
int day;
int year;
} date;

int is_leap_year(int the_year);

int days_in_month(int month,int leap_year);
// Recommendation: use switch statement


void input_date(date *dp);  // User input of form mm/dd/yyyy
                        // where mm, dd and yyy are integers
                        // NO PROMPT  JUST INPUT

void print_date(date d);    // Output format same as input_date's input format

int day_of_year(date d);    // Number of days since beginning of year
                        // January 1 is day 1 (not 0)

int compare_dates(date d1, date d2);
// Returns -1 if d1 earlier than d2, 0 if same date, +1 if d1 after d2

void check_age_and_birthday(date birthdate, date today);
// Output: "You are __ years old and this is your birthday\n"
//      or "You are __ years old and this is not your birthday\n"

void submit_status(date start, date end, date submit);
// Prints ""Submission not accepted - too early\n" if submit < start;
//        "Submission not accepted too - late\n" if end < submit;
//        "Submission accepted\n"  otherwise.
#endif

// Note: A year is a leap year if and only if it is divisible by 4
// and, if it is divisible by 100, it is also divisible by 400.

main.c:

#include "date.h"
#include <stdio.h>

int main()
{
  date d,bdate,today,start,end,submit;

  printf("Enter a date: ");
  input_date(&d);

  printf("\n");
  print_date(d);
  printf(" is day %d of year %d, which is ",day_of_year(d),d.year);
  if (!is_leap_year(d.year))
    printf("not ");
  printf("a leap year\n\n");

  printf("Enter your birthdate: ");
  input_date(&bdate);

  printf("Enter today's date: ");
  input_date(&today);

  printf("\n");
  check_age_and_birthday(bdate,today);

  printf("\nEnter the start date: ");
  input_date(&start);

  printf("Enter the end date: ");
  input_date(&end);

  printf("Enter the date submitted: ");
  input_date(&submit);

  printf("\n\n");
  submit_status(start,end,submit);

  return 0;
}

日期.c:

#include "date.h"

int is_leap_year(int the_year)
{
   if (the_year % 4 == 0) {
                if (the_year % 100 == 0) {
                        if (the_year % 400 == 0) {
                                return (1);
                        }
                        else return (0);
                }
                else return (1);
        }
        else return (0);
}

int days_in_month(int month, int leap_year)
{
        switch (month) {
                case 1: {
                        return (31);
                        break;
                }
                case 2: {
                        if (leap_year) return (29);
                        else return (28);
                        break;
                }
                case 3: {
                        return (31);
                        break;
                }
                case 4: {      
                        return (30);
                        break;
                }
                case 5: {
                        return (31);
                        break;
                }
                case 6: {
                        return (30);
                        break;
                }
                case 7: {
                        return (31);
                        break;
                }
                case 8: {
                        return (31);
                        break;
                }
                case 9: {      
                        return (30);
                        break;
                }
                case 10: {
                        return (31);
                        break;
                }
                case 11: {
                        return (30);
                        break;
                }
                case 12: {
                        return (31);
                        break;
                }
        }              
}

void input_date(date *dp)
{
        do
        {
             scanf("%d/%d/%d", dp->month, dp->day, dp->year);
        }
        while
             (dp->day > days_in_month(dp->month, is_leap_year(dp->year)) || dp->day < 1);
}

void print_date(date d)
{
        printf("%d/%d/%d", d.month, d.day, d.year);
}


int day_of_year(date d)
{
        int day_sum = 0;
        int i;

        if (is_leap_year(d.year))
        {
          for (i = 1; i < d.month; i++)
              {
              day_sum = day_sum + days_in_month(i, 1);
              }

        }
        else
        {
             for (i = 1; i < d.month; i++)
             {
                day_sum += days_in_month(i, 0);
             }
        }

         day_sum += d.day;

         return (day_sum);
}

void check_age_and_birthday(date birthdate, date today)
{
        int age;

        if (today.month > birthdate.month)
        {
                age = today.year - birthdate.year;
        }
        if (today.month < birthdate.month)
        {
                age = (today.year - birthdate.year) - 1;
        }
        if (today.month == birthdate.month)
        {
                if ((today.day > birthdate.day) || (today.day == birthdate.day)) {
                        age = today.year - birthdate.year;
                }
                if (today.day < birthdate.day) {
                        age = (today.year - birthdate.year) - 1;
                }
        }

        if (!compare_dates(birthdate, today))
        {
                printf ("You are %d years old and this is your birthday\n", age);
        }
        else
        {
                printf ("You are %d years old and this is not your birthday\n", age);
        }
}

 int compare_dates(date d1, date d2)
{
        if (d1.year > d2.year)
        {
                return (1);
        }
        if (d1.year < d2.year)
        {
                return (-1);
        }
        if (d1.year == d2.year)
        {
                if (d1.month > d2.month)
                {
                        return (1);
                }
                if (d1.month < d2.month)
                {
                        return (-1);
                }
                if (d1.month == d2.month)
                {
                        if (d1.day > d2.day)
                        {
                                return (1);
                        }
                        if (d1.day < d2.day)
                        {
                                return (-1);
                        }
                        if (d1.day == d2.day)
                        {
                                return (0);
                        }
                }
        }

 }

void submit_status(date start, date end, date submit)
{
        if (compare_dates(submit, start) == -1)
        {
                printf("Submission not accepted - too early\n");
        }
        if (compare_dates(submit, end) == 1)
        {
                printf("Submission not accepted - too late\n");
        }
        if ((compare_dates(submit, start) == 1) && (compare_dates(submit, end) == -1))
        {
                printf("Submission accepted\n");
        }

}

当我将它们放入“控制台应用程序”项目并构建它时，我收到错误:

[Linker error] undefined reference to `submit_status'  
ld returned 1 exit status  
[Build Error] [DateStruct.exe] Error 1

如果我在单独的选项卡中运行它们，并且没有创建项目，我会得到:

mangled line number section.  
[Linker error] undefined reference to `WinMain@16'  
ld returned 1 exit status

我为此使用 Dev-C++。

Answer 1

当您输入日期时，您的程序崩溃，因为您正在传递 dp->month、dp->day 和 dp-> 的值年 到scanf()。 scanf() 需要一个指针来存储结果；在 C 中，整数看起来像一个指针，因此它很高兴地尝试将 dp->month 中的统一值解释为指针，但由于无法在那里写入而崩溃。

您想使用&dp->month等。

我建议在启用所有警告的情况下进行编译(我不知道 Dev-C++，但 GCcflags是 -Wall -Wextra)；我还建议使用 -Werror 处理错误警告，这样您就不会意外地忽略它们。如果我在启用警告的情况下进行编译，GCC 会告诉我有关问题的信息:

cc -Wall -Werror -Wextra    struct.c   -o struct
struct.c: In function ‘main’:
struct.c:13:3: error: format ‘%d’ expects argument of type ‘int *’, but argument 2 has type ‘int’ [-Werror=format]
struct.c:13:3: error: format ‘%d’ expects argument of type ‘int *’, but argument 3 has type ‘int’ [-Werror=format]
cc1: all warnings being treated as errors

这个示例错误也来 self 编写的一个非常快速的测试文件。如果您尝试创建 short, self-contained example，那么在调试问题以及遇到困难时提出问题时，它确实很有帮助。的问题。您发布的代码有很多无关的位；几个头文件、您从未真正接触过的函数等等。将代码分解为问题的最小示例可以帮助您自己找到问题，或者如果失败，请提供一个更容易让其他人浏览并看到问题的示例，而无需费力地完成一堆额外的内容.

这是一个演示该问题的最小示例:

#include <stdio.h>                                                                                      

struct foo {                                                                                            
  int bar;                                                                                              
  int baz;                                                                                              
};                                                                                                      

int main() {                                                                                            
  struct foo x, *px;                                                                                    
  px = &x;                                                                                              

  printf("Enter bar and baz: \n");
  scanf("%d %d", px->bar, px->baz);

  printf("Got: %d %d\n", px->bar, px->baz);

  return 0;
}

以及如何修复它:

#include <stdio.h>                                                                                      

struct foo {                                                                                            
  int bar;                                                                                              
  int baz;                                                                                              
};                                                                                                      

int main() {                                                                                            
  struct foo x, *px;                                                                                    
  px = &x;                                                                                              

  printf("Enter bar and baz: \n");
  scanf("%d %d", &px->bar, &px->baz);

  printf("Got: %d %d\n", px->bar, px->baz);

  return 0;
}