c++ - 为什么 gmtime 和 localtime 给我相同的结果？

Question

我有以下测试代码来查看 gmtime 和 localtime 之间的差异。但他们给了我相同的结果: UTC:2013-05-02T13:59:58 Local:2013-05-02T13:59:58

time_t now;
time(&now);
tm *pTimeStruct = gmtime(&now);
tm *plocalTimeStruct = localtime(&now);

string timeStr = "";
char timeBuf[64] = {'\0'};

sprintf(timeBuf,"UTC:%-4.4d-%-2.2d-%-2.2dT%-2.2d:%-2.2d:%-2.2d "
    "Local:%-4.4d-%-2.2d-%-2.2dT%-2.2d:%-2.2d:%-2.2d",
    (pTimeStruct->tm_year + 1900),
    (pTimeStruct->tm_mon + 1),
    pTimeStruct->tm_mday,
    pTimeStruct->tm_hour,
    pTimeStruct->tm_min,
    pTimeStruct->tm_sec,
    (plocalTimeStruct->tm_year + 1900),
    (plocalTimeStruct->tm_mon + 1),
    plocalTimeStruct->tm_mday,
    plocalTimeStruct->tm_hour,
    plocalTimeStruct->tm_min,
    plocalTimeStruct->tm_sec);

timeStr += timeBuf;
cout << timeStr << endl;

编辑:

我在东部时区。

编辑2:

更新的代码使用 diff 结构，但得到相同的结果:

        time_t now;
        time(&now);
        time_t now2;
        time(&now2);
        tm *pTimeStruct = gmtime(&now);
        tm *plocalTimeStruct = localtime(&now2);

Answer 1

您需要复制对 gmtime 和 localtime 的调用之间的值:

The return value points to a statically allocated struct which might be overwritten by subsequent calls to any of the date and time functions.

说我系统上的手册页。至少在 Linux 上这是常见行为。