std::chrono::d uration 可以按秒初始化，但不能按毫秒初始化?

Question

作品：std::chrono::duration<unsigned long long> test1 = std::chrono::seconds(1);

不起作用：std::chrono::duration<unsigned long long> test2 = std::chrono::milliseconds(1);

为什么会有差异？持续时间在内部是否没有足够的粒度？

从以毫秒为单位的值初始化持续时间的首选方法是什么？

Answer 1

std::chrono::duration的模板参数列表有两个参数：保存基础数据的类型和表示持续时间指数的std::ratio参数。std::chrono::seconds和std::chrono::milliseconds等类型是该模板的专用化，分别使用std::chrono::duration<int64_t, std::ratio<1>>和std::chrono::duration<int64_t, std::ratio<1, 1000>>。

如果不为类型提供std::ratio参数，它将默认为std::ratio<1>。

因此，您的自定义持续时间类型隐式采用std::chrono::duration<unsigned long long, std::ratio<1>>形式，这使得它几乎等同于std::chrono::seconds(唯一的区别是无符号值而不是有符号值(，但由于其比率高于提供给std::chrono::milliseconds的比率，类模板禁止赋值/原始转换。对于这种情况，如果您希望分配通过，则需要显式强制转换它：

typedef std::duration<unsigned long long> my_duration;
//my_duration test1 = std::chrono::milliseconds(1);//Forbidden
my_duration test1 = std::chrono::duration_cast<my_duration>(std::chrono::milliseconds(1)); //Permitted, will be truncated
my_duration test2 = std::chrono::duration_cast<my_duration>(1ms); //Permitted, may be easier to read

std::ratio参数表示持续时间的即时报价大小。即时报价大小越小，代表持续时间的基础数字就越大。举个例子：

using seconds = std::chrono::seconds; //std::duration<int64_t, std::ratio<1,1>>
using milliseconds = std::chrono::milliseconds; //std::duration<int64_t, std::ratio<1,1000>>
using nanoseconds = std::chrono::nanoseconds; //std::duration<int64_t, std::ratio<1,1000000000>>
seconds time1 = 5s; //underlying integer value is 5.
milliseconds time2 = 5ms; //underlying integer value is 5.
time2 = time1; //underlying integer value for time2 is 5000, for time1 is still 5.
time2 = 17ms; //underlying integer value is 17.
//time1 = time2; //forbidden; tick size of time2 is smaller than time1
time1 = std::chrono::duration_cast<seconds>(time2); 
//underlying integer value for time1 is 0, due to truncation; time2 is still 17.
nanoseconds time3 = 5ms; //underlying integer value is 5000000.
time1 = 2s; //underlying integer value is 2.
time3 = time1; //underlying integer value is 2000000000.