Chrono

Question

如何获得（使用 std::chrono 库）两个时间点之间的毫秒差？

我可以用这个来做到这一点：

std::chrono::time_point<std::chrono::system_clock> now = std::chrono::system_clock::now();

std::chrono::time_point<std::chrono::system_clock> foo = now + std::chrono::milliseconds(100);

std::chrono::duration<float> difference = foo - now;

const int milliseconds = difference.count() * 1000;

我怎样才能得到这个时间以毫秒为单位，所以我可以将持续时间用作无符号整数，而不是浮点数，然后乘以 1000？

Answer 1

http://www.cplusplus.com/reference/chrono/duration_cast/

std::chrono::duration_cast<std::chrono::milliseconds>();

Answer 2

chrono::duration_cast<chrono::milliseconds>(end_time - start_time).count()

Answer 3

std::chrono::duration has two template parameters, the second being exactly the unit of measure. You can invoke std::chrono::duration_cast to cast from one duration type to another. Also, there is a predefined duration type for milliseconds: std::chrono::milliseconds。组合在一起：

auto milliseconds = std::chrono::duration_cast<std::chrono::milliseconds>(foo - now);

要获得实际的毫秒数，请使用 duration::count:

auto ms = milliseconds.count();

它的 return 类型是 duration::rep，对于像 std::chrono::milliseconds 这样的标准持续时间类型，它是一个未指定大小的有符号整数。

Answer 4

我在使用 e-09 等字母打印持续时间时遇到问题。这是我修复它的方法：

auto start = std::chrono::high_resolution_clock::now();
< run your function or code here >
auto finish = std::chrono::high_resolution_clock::now();
std::chrono::duration<double, std::milli> elapsed = finish - start;
std::cout << "Elapsed Time: " << elapsed.count() << " seconds" << std::endl;

现在我得到了想要的结果：

Elapsed Time: 34.406 seconds

Chrono - 两个时间点之间的差异（以毫秒为单位）？

Chrono - The difference between two points in time in milliseconds?

c++

time

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