c - 尝试在 struct tm 中设置值

Question

我正在尝试创建一个函数来获取当前日期、获取个人生日并计算他们的年龄。我得到了当前日期。现在我尝试获取该人的生日(通过 individual 传递给函数)并将该人的生日值存储在名为 str_bday 的 struct tm 变量中。该程序编译，但当我运行它时，我得到这个:

       Your last name is Smith
       Sat Feb  8 16:04:05 2014
       Your birthday is:  3/1/1940
       Wed Dec 31 18:59:59 1969
       v245-2%

我不明白为什么当我打印出 str_bday struct tm 变量时，它会打印出他们的生日是 1969 年。有人能帮帮我吗。以下是我的此功能的代码:

       char* calcage(char *individual, char *age)
         {

           time_t current_time;
           char *c_time_string;

           current_time = time(NULL);


           c_time_string = ctime(&current_time);

           printf(c_time_string);



           char *birthday = (char *)malloc(50*sizeof(char));
           birthday = strrchr(individual, ',');
           birthday++;

           printf("Your birthday is:  %s\n", birthday);


           char *bmonth, *bday, *byear;
           int numbmonth, numbday,  numbyear;


           bmonth = strtok(birthday, "/");
           bday = strtok(NULL, "/");
           byear = strtok(NULL, "/");

           numbmonth = atoi(bmonth);
           numbday = atoi(bday);
           numbyear = atoi(byear);


           struct tm str_bday;
           time_t time_bday;

           str_bday.tm_year = 2012;

           time_bday = mktime(&str_bday);
           printf(ctime(&time_bday));



         }

Answer 1

tm_year 将根据 1900 设置年份。 (例如str_bday.tm_year = numbyear -1900;)

还应按如下方式检查mktime的返回值。

if(time_bday == (time_t)-1)
    printf("error");

如果早于一年，它会返回错误，因为它在许多系统上基于 1970 年。

如果是这样的话，必须自当日起进行处理。

<小时/>

测试代码

#include <stdio.h>
#include <time.h>
int main(){
    struct tm str_bday = {0};
    time_t time_bday;

    str_bday.tm_year = 2012 - 1900;
    str_bday.tm_mon = 3 - 1;
    str_bday.tm_mday = 1;
    time_bday = mktime(&str_bday);
    if(time_bday == (time_t)-1)
        printf("error\n");
    else
        printf("%s\n", ctime(&time_bday));//Thu Mar 01 00:00:00 2012
    return 0;
}