C++ C 中从 Unix 纪元时间转换的损坏

至少是这些问题:

Not checking the return value个

gmtime()和time()可能返回NULL、-1，指示失败.健壮的代码测试这一点.

一百零二

每次调用gmtime()和localtime()都会返回指向相同struct tm的指针.当需要保留之前的值时，保存内容.

重写示例

int main(void) {
    long long timestamp_ms = 1692457488113;
    char formatted_time[50]; // only need 20 characters to hold "MM/DD/YYYY HH:MM:SS"

    time_t timestamp_seconds = (time_t)(timestamp_ms / 1000); // Convert milliseconds to seconds
    struct tm *timestamp_timeinfo = gmtime(&timestamp_seconds); // Use gmtime for UTC time
    if (timestamp_timeinfo == NULL) {
      fprintf(stderr, "Failure to convert\n");
      return -1;
    }
    struct tm t1 = *timestamp_timeinfo;

    // print result - should be 08/19/2023 15:04:48  (Unix Epoch time: 1692457488)
    if (0 == strftime(formatted_time, sizeof(formatted_time), "%m/%d/%Y %H:%M:%S", &t1)) {
      fprintf(stderr, "Failure strtime\n");
      return -1;
    }
    printf("1st Printing of Timestamp (UTC): %s  (Unix Epoch time: %lld)\n", formatted_time, timestamp_ms/1000);

       
    time_t current_utc_time; // Get current UTC time
    if (time(&current_utc_time) == -1) {
      fprintf(stderr, "Failure to convert\n");
      return -1;
    }
    struct tm *current_timeinfo = gmtime(&current_utc_time);  
    if (current_timeinfo == NULL) {
      fprintf(stderr, "Failure to convert\n");
      return -1;
    }

    // print again
    strftime(formatted_time, sizeof(formatted_time), "%m/%d/%Y %H:%M:%S", &t1);
    printf("2nd Printing of Timestamp (UTC): %s  (Unix Epoch time: %lld)\n", formatted_time, timestamp_ms/1000);

    return 0;
} 

次要问题:

char formatted_time[50]; // only need 20 characters to hold "MM/DD/YYYY HH:MM:SS"是危险的，行程代码判断器，因为访问的struct tm个成员是int，他们may的值只要是文本版本INT_MIN.如果年份是5位数呢？strftime()不假定成员在其主要范围内，代码的其余部分也不应假定.最好根据格式调整大小的缓冲区.

与其在缓冲空间上吝啬，不如慷慨一些.

#define TIME_FMT "%m/%d/%Y %H:%M:%S"
#define INT_LEN 11
#define FTIME_SIZE (6 * INT_LEN + sizeof(TIME_FMT))

char formatted_time[FTIME_SIZE];
strftime(formatted_time, sizeof(formatted_time), TIME_FMT, &t1);