c - 由单独的本地时间调用的两个 struct tm info 返回相同的值

Question

我有这个测试代码:

  1 #include <stdio.h>
  2 #include <time.h>
  3 
  4 int main() {
  5     struct tm *info1;                                                                                                                                      
  6     struct tm *info2;
  7     unsigned long i = 100000000;
  8     unsigned long j = 200000000;
  9     
 10     info1 = localtime((time_t *) &i);
 11     info2 = localtime((time_t *) &j);
 12     
 13     printf("%s(): info1->tm_sec = %d\n", __func__, info1->tm_sec);
 14     printf("%s(): info1->tm_min = %d\n", __func__, info1->tm_min);
 15     printf("%s(): info1->tm_hour = %d\n", __func__, info1->tm_hour);
 16     printf("%s(): info1->tm_mday = %d\n", __func__, info1->tm_mday);
 17     printf("%s(): info1->tm_mon = %d\n", __func__, info1->tm_mon);
 18     printf("%s(): info1->tm_year = %d\n", __func__, info1->tm_year);
 19     
 20     printf("%s(): info2->tm_sec = %d\n", __func__, info2->tm_sec);
 21     printf("%s(): info2->tm_min = %d\n", __func__, info2->tm_min);
 22     printf("%s(): info2->tm_hour = %d\n", __func__, info2->tm_hour);
 23     printf("%s(): info2->tm_mday = %d\n", __func__, info2->tm_mday);
 24     printf("%s(): info2->tm_mon = %d\n", __func__, info2->tm_mon);
 25     printf("%s(): info2->tm_year = %d\n", __func__, info2->tm_year);
 26     
 27     
 28 
 29     return 0;
 30 }

输出是:

main(): info1->tm_sec = 20
main(): info1->tm_min = 33
main(): info1->tm_hour = 3
main(): info1->tm_mday = 4
main(): info1->tm_mon = 4
main(): info1->tm_year = 76
main(): info2->tm_sec = 20
main(): info2->tm_min = 33
main(): info2->tm_hour = 3
main(): info2->tm_mday = 4
main(): info2->tm_mon = 4
main(): info2->tm_year = 76

第 7 行和第 8 行实际上是时间戳(自 Epoch 以来的秒数)，作为从调用函数传递的 unsigned long(我只是在这里硬编码)。

第 10 行和第 11 行是我关心的问题。我需要获取两个时间戳 i 和 j 的 struct tm 信息。基本上，我需要获取 info1 的月份并将其与 info2 的月份等进行比较。

在第 13 到 25 行打印，info1 和 info2 返回相同的值(即相同的秒数、相同的分钟数、相同的小时数等)。

两个问题:

1. 为什么他们有相同的值(value)观？
2. 如何获取info1和info2的不同值？

Answer 1

localtime() -> 它的返回值指向一个静态分配的结构，该结构可能会被后续调用任何日期和时间函数覆盖。所以使用 localtime_r() 将数据存储在用户提供的结构中。

#include <stdio.h>
   #include <time.h>

   int main() {
      struct tm info1;                                                                                                                                      
      struct tm info2;
     unsigned long i = 100000000;
      unsigned long j = 200000000;

     localtime_r((time_t *) &i,&info1);
       localtime_r((time_t *) &j,&info2);

      printf("%s(): info1->tm_sec = %d\n", __func__, info1.tm_sec);
      printf("%s(): info1->tm_min = %d\n", __func__, info1.tm_min);
      printf("%s(): info1->tm_hour = %d\n", __func__, info1.tm_hour);
     printf("%s(): info1->tm_mday = %d\n", __func__, info1.tm_mday);
      printf("%s(): info1->tm_mon = %d\n", __func__, info1.tm_mon);
      printf("%s(): info1->tm_year = %d\n", __func__, info1.tm_year);

      printf("%s(): info2->tm_sec = %d\n", __func__, info2.tm_sec);
     printf("%s(): info2->tm_min = %d\n", __func__, info2.tm_min);
      printf("%s(): info2->tm_hour = %d\n", __func__, info2.tm_hour);
      printf("%s(): info2->tm_mday = %d\n", __func__, info2.tm_mday);
printf("%s(): info2->tm_mon = %d\n", __func__, info2.tm_mon);
     printf("%s(): info2->tm_year = %d\n", __func__, info2.tm_year);



      return 0;
  }