我构建了一个函数,它将特定日期作为输入并以 std::chrono::milliseconds
格式返回该日期。
milliseconds lowerRangeBound = TimeStamp(mm, dd, HH, MM, SS, yyyy);
例如,
milliseconds a = TimeStamp(8/*month*/, 23/*day*/, 14/*hours*/, 46/*minutes*/, 32/*seconds*/, 2017/*year*/);
以转换后的字符串格式返回:2017.08.23-14.46.32
我现在想做但行不通的是给定两个日期 ( milliseconds
) 以在这两个日期定义的范围内取一个随机日期。例如,给定
milliseconds a = TimeStamp(8/*month*/, 23/*day*/, 13/*hours*/, 46/*minutes*/, 32/*seconds*/, 2017/*year*/);
milliseconds b = TimeStamp(10/*month*/, 23/*day*/, 13/*hours*/, 46/*minutes*/, 32/*seconds*/, 2017/*year*/);
预期的输出是一个 milliseconds c
,它在字符串格式中是一个像这样的日期 2017.09.13-12.56.12
。请注意,所需的输出是 milliseconds
,提供字符串格式是为了在 readable format 中说话。
到目前为止我尝试的是将每个 milliseconds
变量转换为 long
数字 ( .count()
) 并获得 long
范围内的 random [a,b]
。然而,输出日期是无关紧要的:1977.12.06-16.27.02
。
你能帮帮我吗?
提前谢谢你。
编辑:下面的代码是受此 link 的启发
milliseconds TimeStamp(int mm, int dd, int HH, int MM, int SS, int yyyy) {
tm ttm = tm();
ttm.tm_year = yyyy - 1900; // Year since 1900
ttm.tm_mon = mm - 1; // Month since January
ttm.tm_mday = dd; // Day of the month [1-31]
ttm.tm_hour = HH; // Hour of the day [00-23]
ttm.tm_min = MM;
ttm.tm_sec = SS;
time_t ttime_t = mktime(&ttm);
system_clock::time_point time_point_result = std::chrono::system_clock::from_time_t(ttime_t);
milliseconds now_ms = std::chrono::time_point_cast<std::chrono::milliseconds>(time_point_result).time_since_epoch();
return now_ms;
}
milliseconds getRandomTimestamp(int mm_1, int dd_1, int HH_1, int MM_1, int SS_1, int yyyy_1,
int mm_2, int dd_2, int HH_2, int MM_2, int SS_2, int yyyy_2, int N) {
milliseconds lowerRangeBound = fixedTimeStamp(mm_1, dd_1, HH_1, MM_1, SS_1, yyyy_1);
milliseconds upperRangeBound = fixedTimeStamp(mm_2, dd_2, HH_2, MM_2, SS_2, yyyy_2);
long lowerRange_ = lowerRangeBound.count();
long upperRange_ = upperRangeBound.count();
//long output = rand() % (upperRange_ - lowerRange_ + 1) + lowerRange_;
// rand() replaced after @Jarod42's suggestion.
std::default_random_engine generator;
std::uniform_int_distribution<int> distribution(lowerRange_, upperRange_);
long output = distribution(generator);
std::chrono::duration<long> dur(output);
return dur;
}
最佳答案
可能是因为混合问题,您对数学感到困惑。
我建议您以毫秒为单位编写“随机时间戳”例程。
您可以在单独的步骤中执行从/到时间戳的转换。
我会这样写例程:
#include<random>
#include<chrono>
#include<cassert>
#include<iostream>
template<class Eng>
std::chrono::milliseconds getRandomTimestamp(Eng& eng,
std::chrono::milliseconds low,
std::chrono::milliseconds high)
{
// deduce the actual integer type.
// Otherwise we'll have to guess what T is when using
// uniform_int_distribution<T>
using count_type = decltype(low.count());
// from this we can deduce the correct integer distribution
using dist_type = std::uniform_int_distribution<count_type>;
// create the distribution generator
auto dist = dist_type(low.count(), high.count());
// and apply it to the random generator, returning
// milliseconds somewhere between low and high
return std::chrono::milliseconds(dist(eng));
}
int main()
{
using namespace std::literals;
std::random_device dev;
std::default_random_engine eng(dev());
auto t1 = 10'000'000ms;
auto t2 = 11'000'000ms;
auto t3 = getRandomTimestamp(eng, t1, t2);
assert(t1 <= t3);
assert(t3 <= t2);
std::cout << t1.count() << std::endl;
std::cout << t2.count() << std::endl;
std::cout << t3.count() << std::endl;
}
关于c++ - 我们可以从相同类型的范围中获取 `std::chrono::milliseconds` 变量吗?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/46825247/