使用 ctime 和 time_t 的输出差异

Question

在处理时间和持续时间时发现了一个奇怪的差异，通过 ctime 输出 time_t。我绝对在这里遗漏了一些东西。在代码和输出之后有更多关于问题的信息。我搜索了一下类似的问题，如果有请指导我。

我目前正在运行： 操作系统：Windows 7 64 位 CPU：英特尔酷睿 i5 M 520 2.40Ghz 内存：4GB IDE：Microsoft Visual Studio Express 2013

编辑：完整来源

#include <iostream>
#include <chrono>
#include <ctime>
#include <thread>

using namespace std::chrono;

int main()
{
    duration<int, std::milli> wait(5500);

    time_point<system_clock>  timePoint_Start;
    time_point<system_clock>  timePoint_End;
    time_t                    timeT_Start;
    time_t                    timeT_End;

    std::cout << "01. timePoint_Start Address, sizeof: " << &timePoint_Start << ", " << sizeof(timePoint_Start) << std::endl; 
    std::cout << "02. timePoint_End   Address, sizeof: " << &timePoint_End   << ", " << sizeof(timePoint_End)   << std::endl;
    std::cout << "03. timeT_Start     Address, sizeof: " << &timeT_Start     << ", " << sizeof(timeT_Start)     << std::endl;
    std::cout << "04. timeT_End       Address, sizeof: " << &timeT_End       << ", " << sizeof(timeT_End)       << std::endl;

    timePoint_Start = system_clock::now();     // < = = = = set the time the 1st time
    timeT_Start     = system_clock::to_time_t(timePoint_Start);
    std::cout << "05. Time Start:   " << system_clock::to_time_t(timePoint_Start) << " " << std::ctime(&timeT_Start);

    std::this_thread::sleep_for(wait);

    timePoint_End   = system_clock::now();     // < = = = = set the time the 2nd time
    timeT_End       = system_clock::to_time_t(timePoint_End);
    std::cout << "06. Time End:     " << system_clock::to_time_t(timePoint_End)   << " " << std::ctime(&timeT_End);

    duration<double> elapsed_seconds = timePoint_End - timePoint_Start;

    std::cout << "07. Time Start:   " << system_clock::to_time_t(timePoint_Start)    << " 0x" << &timeT_Start << " " << std::ctime(&timeT_Start)
              << "08. Time Wait:    duration<int, std::milli> wait(5500);"           << std::endl
              << "09. Time End:     " << system_clock::to_time_t(timePoint_End)      << " 0x" << &timeT_End   << " " << std::ctime(&timeT_End)
              << "10. Time Elapsed: " << elapsed_seconds.count() << "s"              << std::endl;

    std::cout << "11. Time End:     " << system_clock::to_time_t(timePoint_End)      << " 0x" << &timeT_End   << " " << std::ctime(&timeT_End);

这是输出中出现差异的地方编辑：完整输出

01. timePoint_Start Address, sizeof: 0046FAA4, 8
02. timePoint_End   Address, sizeof: 0046FA94, 8
03. timeT_Start     Address, sizeof: 0046FA84, 8
04. timeT_End       Address, sizeof: 0046FA74, 8
05. Time Start:   1402880407 Sun Jun 15 20:00:07 2014
06. Time End:     1402880413 Sun Jun 15 20:00:13 2014
07. Time Start:   1402880407 0x0046FA84 Sun Jun 15 20:00:07 2014
08. Time Wait:    duration<int, std::milli> wait(1500);
09. Time End:     1402880413 0x0046FA74 Sun Jun 15 20:00:07 2014
10. Time Elapsed: 5.50932s
11. Time End:     1402880413 0x0046FA74 Sun Jun 15 20:00:13 2014

据我了解，输出的第 6 行、第 9 行和第 11 行的时间戳应该匹配。输出的第 9 行似乎选择了错误的引用。这里有什么语法错误吗？

我是一个菜鸟，但这看起来真的很奇怪。如果我在错误的部分，请引导我到正确的部分。真的很想知道这里发生了什么。

Answer 1

std::ctime() 函数来自 C 标准，我相信，在 C 标准中，它返回一个指向静态内存位置的指针。这一次只能存储一个值。由于您有一个使用 std::ctime() 3 次的巨大 std::cout << ... 调用，因此您很不走运；您将在所有三个位置显示一个值，而不是 3 个不同的值（通常 — 尽管细节取决于对 ctime() 的调用如何与对 << 运算符的调用交错，这取决于编译器）。

证明这一点的最简单方法是修改代码，使标记为 7、8、9、10 的行成为单独的语句。

长期修复需要更多思考。