c++ - 在cout中进行算术时，chrono duration_cast不起作用

我不了解以下行为

unsigned long begin_time = \
    std::chrono::duration_cast<std::chrono::milliseconds>(std::chrono::steady_clock::now().time_since_epoch()).count();

//some code here

std::cout << "time diff with arithmetic in cout: " << \
    std::chrono::duration_cast<std::chrono::milliseconds>(std::chrono::steady_clock::now().time_since_epoch()).count() - begin_time << std::endl;

unsigned long time_diff = \
    std::chrono::duration_cast<std::chrono::milliseconds>(std::chrono::steady_clock::now().time_since_epoch()).count() - begin_time;

std::cout << "time_diff: " << time_diff << std::endl;

输出:

time diff with arithmetic in cout: <very large number (definitely not milliseconds)>
time_diff: <smaller number (definitely milliseconds)>

当我在cout中进行算术运算时，为什么duration_cast不起作用？我已经将unsigned int和int用于time_diff变量，但是当我第一次在变量初始化或赋值内进行算术运算时，始终可以获得良好的输出。

注意

我正在使用Visual Studio 2013(社区版)

最佳答案

您可能正在溢出unsigned long(sizeof为4):

unsigned long begin_time = \
    std::chrono::duration_cast<std::chrono::milliseconds>(std::chrono::steady_clock::now().time_since_epoch()).count();

推荐的:

using namespace std::chrono;
auto begin_time = steady_clock::now();

//some code here

std::cout << "time diff with arithmetic in cout: " <<
    duration_cast<milliseconds>(steady_clock::now() - begin_time).count() << std::endl;