Question

我正在尝试创建一个获取当前日期，接受个人生日并计算其年龄的函数。我得到了当前的日期。现在我试图接受这个人的生日（通过个人传递给该功能）并将该人的生日值存储在名为str_bday的结构tm变量中。程序编译，但是当我运行它时，我得到了这个：

       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%

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

       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;）

另外，您应该按照以下mktime的返回值进行检查。

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

如果在一年前返回错误，因为它基于1970年的许多系统。

必须自己处理，如果是这样的话。

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;
}

尝试在struct tm中设置值

1 个答案: