我为大学班级编写了三种算法版本。
一个是蛮力,另一个是贪婪,最后一个是启发式。
我希望能够衡量每个算法需要多少时间才能完成。
我正在使用<chrono>
库来实现这一目标
现在,我的代码如下所示:
#include <iostream>
#include <chrono>
#include <sstream>
using namespace std;
string getTimeElapsed(long time1, const string &unit1, long time2 = 0, const string &unit2 = "") {
stringstream s;
s << time1 << " [" << unit1 << "]";
if (time2) s << " " << time2 << " [" << unit2 << "]";
return s.str();
}
int main() {
auto begin = chrono::system_clock::now();
// algorithm goes here
auto solution = /* can be anything */
auto end = chrono::system_clock::now();
auto diff = end - begin;
string timeElapsed;
auto hours = chrono::duration_cast<chrono::hours>(diff).count();
auto minutes = chrono::duration_cast<chrono::minutes>(diff).count();
if (hours) {
minutes %= 60;
timeElapsed = getTimeElapsed(hours, "h", minutes, "min");
} else {
auto seconds = chrono::duration_cast<chrono::seconds>(diff).count();
if (minutes) {
seconds %= 60;
timeElapsed = getTimeElapsed(minutes, "min", seconds, "s");
} else {
auto milliseconds = chrono::duration_cast<chrono::milliseconds>(diff).count();
if (seconds) {
milliseconds %= 1000;
timeElapsed = getTimeElapsed(seconds, "s", milliseconds, "ms");
} else {
auto microseconds = chrono::duration_cast<chrono::microseconds>(diff).count();
if (milliseconds) {
microseconds %= 1000;
timeElapsed = getTimeElapsed(milliseconds, "ms", microseconds, "μs");
} else {
auto nanoseconds = chrono::duration_cast<chrono::nanoseconds>(diff).count();
if (microseconds) {
nanoseconds %= 1000;
timeElapsed = timeElapsed = getTimeElapsed(microseconds, "μs", nanoseconds, "ns");
} else timeElapsed = getTimeElapsed(nanoseconds, "ns");
}
}
}
}
cout << "Solution [" << solution << "] found in " << timeElapsed << endl;
return 0;
}
如您所见,堆叠的
if-else
子句看起来非常丑陋,您可以在此处看到一个模式:if (timeUnit) {
timeElapsed = /* process current time units */
} else {
/* step down a level and do the same for smaller time units */
}
我想使该过程成为递归函数。
但是,我不知道该函数的参数是什么,因为
chrono::duration
是模板结构(?)这个函数看起来像这样:
string prettyTimeElapsed(diff, timeUnit) {
// recursion bound condition
if (timeUnit is chrono::nanoseconds) return getTimeElapsed(timeUnit, "ns");
auto smallerTimeUnit = /* calculate smaller unit using current unit */
if (timeUnit) return getTimeElapsed(timeUnit, ???, smallerTimeUnit, ???);
else return prettyTimeElapsed(diff, smallerTimeUnit);
}
我正在考虑这样做:
auto timeUnits = {chrono::hours(), chrono::minutes(), ..., chrono::nanoseconds()};
然后,我可以将指针(甚至索引)带到时间单位,并将其传递给函数。
问题是我不知道如何概括这些结构。
CLion突出显示错误
Deduced conflicting types (duration<[...], ratio<3600, [...]>> vs duration<[...], ratio<60, [...]>>) for initializer list element type
最佳答案
使用chrono
时,最好的一般建议是仅在绝对必要时才转义类型系统(使用.count()
)。这可能与C或某些不了解chrono的C ++库接口。在C ++ 20之前,这还意味着输出到流。
如果我们将自己保留在类型系统中,则可以得到很多正确的转换。
让我们更正问题中的代码以反映这一点:
#include <iostream>
#include <chrono>
#include <sstream>
std::string getTimeElapsed(long time1, const std::string &unit1, long time2 = 0, const std::string &unit2 = "") {
std::stringstream s;
s << time1 << " [" << unit1 << "]";
if (time2) s << " " << time2 << " [" << unit2 << "]";
return s.str();
}
int main() {
auto begin = std::chrono::system_clock::now();
// algorithm goes here
auto solution = "solution"; /* can be anything */
auto end = std::chrono::system_clock::now();
auto diff = end - begin;
std::string timeElapsed{""};
// Let's make the typing and reading easier for us but requires C++14
using namespace std::chrono_literals;
auto hours = std::chrono::duration_cast<std::chrono::hours>(diff);
auto minutes = std::chrono::duration_cast<std::chrono::minutes>(diff % 1h);
if (hours != 0h) {
// We need to escape the type system to call getTimeElapsed
timeElapsed = getTimeElapsed(hours.count(), "h", minutes.count(), "min");
} else {
auto seconds = std::chrono::duration_cast<std::chrono::seconds>(diff % 1min);
if (minutes != 0min) {
timeElapsed = getTimeElapsed(minutes.count(), "min", seconds.count(), "s");
} else {
auto milliseconds = std::chrono::duration_cast<std::chrono::milliseconds>(diff % 1s);
if (seconds != 0s) {
timeElapsed = getTimeElapsed(seconds.count(), "s", milliseconds.count(), "ms");
} else {
auto microseconds = std::chrono::duration_cast<std::chrono::microseconds>(diff % 1ms);
if (milliseconds != 0ms) {
timeElapsed = getTimeElapsed(milliseconds.count(), "ms", microseconds.count(), "μs");
} else {
auto nanoseconds = std::chrono::duration_cast<std::chrono::nanoseconds>(diff % 1us);
if (microseconds != 0us) {
timeElapsed = timeElapsed = getTimeElapsed(microseconds.count(), "μs", nanoseconds.count(), "ns");
} else timeElapsed = getTimeElapsed(nanoseconds.count(), "ns");
}
}
}
}
std::cout << "Solution [" << solution << "] found in " << timeElapsed << std::endl;
return 0;
}
现在,我们一直坚持使用
chrono
。调用getTimeElapsed
尚不兼容。我并不完全满意,因此我们也支持
chrono
中的duration
:template <typename Duration1, typename Duration2>
std::string getTimeElapsed(Duration1 time1, const std::string &unit1, Duration2 time2, const std::string &unit2) {
std::stringstream s;
s << time1.count() << " [" << unit1 << "]";
if (time2 != Duration2::zero()) s << " " << time2.count() << " [" << unit2 << "]";
return s.str();
}
template <typename Duration1>
std::string getTimeElapsed(Duration1 time1, const std::string &unit1) {
std::stringstream s;
s << time1.count() << " [" << unit1 << "]";
return s.str();
}
我们需要两个版本的
getTimeElapsed
,因为在最后一个getTimeElapsed
中,我们仅使用一个时间和单位参数对,这意味着我们不能满足两种else
类型的template
参数要求。现在,代码看起来好多了(仅保留相关更改):
...
if (hours != 0h) {
timeElapsed = getTimeElapsed(hours, "h", minutes, "min");
} else {
auto seconds = std::chrono::duration_cast<std::chrono::seconds>(diff % 1min);
if (minutes != 0min) {
timeElapsed = getTimeElapsed(minutes, "min", seconds, "s");
} else {
auto milliseconds = std::chrono::duration_cast<std::chrono::milliseconds>(diff % 1s);
if (seconds != 0s) {
timeElapsed = getTimeElapsed(seconds, "s", milliseconds, "ms");
} else {
auto microseconds = std::chrono::duration_cast<std::chrono::microseconds>(diff % 1ms);
if (milliseconds != 0ms) {
timeElapsed = getTimeElapsed(milliseconds, "ms", microseconds, "μs");
} else {
auto nanoseconds = std::chrono::duration_cast<std::chrono::nanoseconds>(diff % 1us);
if (microseconds != 0us) {
timeElapsed = timeElapsed = getTimeElapsed(microseconds, "μs", nanoseconds, "ns");
} else timeElapsed = getTimeElapsed(nanoseconds, "ns");
}
}
}
}
...
大!但是,我们仍然邀请用户将他们想要的任何内容发送到
Duration
,除非他们碰巧拥有getTimeElapsed
成员,否则将导致编译器错误。让我们限制一下
.count()
:template <typename Rep1, typename Ratio1, typename Rep2, typename Ratio2>
std::string getTimeElapsed(std::chrono::duration<Rep1, Ratio1> time1, const std::string &unit1, std::chrono::duration<Rep2, Ratio2> time2, const std::string &unit2) {
std::stringstream s;
s << time1.count() << " [" << unit1 << "]";
if (time2 != time2.zero()) s << " " << time2.count() << " [" << unit2 << "]";
return s.str();
}
template <typename Rep, typename Ratio>
std::string getTimeElapsed(std::chrono::duration<Rep, Ratio> time1, const std::string &unit1) {
std::stringstream s;
s << time1.count() << " [" << unit1 << "]";
return s.str();
}
我们不需要更改此代码。我相信这足以帮助您了解如何在更通用的上下文中使用
template
,这是您遇到的一个子问题。现在我们可以开始解决您的问题了,我认为(通过阅读评论)实际上是“我如何整理嵌套的
std::chrono::duration
语句并仅打印前两个非零单位”。这并不像它第一次出现那样简单。我认为,递归很少是答案。将其视为单元类型的循环也是对其的过度设计,您将需要编写一些代码来从元组中获取当前类型的索引,将其增加一个,然后使用该代码为相同的元组建立索引以获得更高分辨率的下一个单元。然后,当所有事情都说完之后,您仍然需要知道要打印哪个单元才能为该值提供上下文。我宁愿看到
if
编写如下:std::string getTimeElapsed(std::chrono::nanoseconds elapsed, size_t maxUnits = 2)
{
using namespace std::chrono_literals;
std::ostringstream formatted("");
int usedUnits{};
auto hours = std::chrono::duration_cast<std::chrono::hours>(elapsed);
if (hours != 0h)
{
formatted << hours.count() << " [h] ";
++usedUnits;
}
auto minutes = std::chrono::duration_cast<std::chrono::minutes>(elapsed % 1h);
if (minutes != 0min)
{
formatted << minutes.count() << " [min] ";
++usedUnits;
}
auto seconds = std::chrono::duration_cast<std::chrono::seconds>(elapsed % 1min);
if (seconds != 0min && usedUnits != maxUnits)
{
formatted << seconds.count() << " [s] ";
++usedUnits;
}
auto milliseconds = std::chrono::duration_cast<std::chrono::milliseconds>(elapsed % 1s);
if (milliseconds != 0ms && usedUnits != maxUnits)
{
formatted << milliseconds.count() << " [ms] ";
++usedUnits;
}
auto microseconds = std::chrono::duration_cast<std::chrono::microseconds>(elapsed % 1ms);
if (microseconds != 0us && usedUnits != maxUnits)
{
formatted << microseconds.count() << " [us] ";
++usedUnits;
}
auto nanoseconds = std::chrono::duration_cast<std::chrono::nanoseconds>(elapsed % 1us);
if (nanoseconds != 0us && usedUnits != maxUnits)
{
formatted << nanoseconds.count() << " [us] ";
++usedUnits;
}
return formatted.str();
}
将经过的总时间作为
getTimeElapsed
(您已经在std::chrono::nanoseconds
中获得),并将其传递给end - begin
。现在,我们进行与以前相同的计算以获得组件单位,但也跟踪我们已经计算了多少个单位。如果getTimeElapsed
为1'000'000'000ns,则结果为“ 1 [s]”;如果elapsed
为1'234'568ns,则结果为“ 1 [ms] 234 [us]”。有尾随的空间,但我将留给您解决。这也意味着我们不再需要之前重构过的
elapsed
,但是我添加了它们以显示我在整个重构过程中的思考过程。最终程序如下所示:#include <chrono>
#include <iostream>
#include <sstream>
std::string getTimeElapsed(std::chrono::nanoseconds elapsed, size_t maxUnits = 2)
{
using namespace std::chrono_literals;
std::ostringstream formatted("");
int usedUnits{};
auto hours = std::chrono::duration_cast<std::chrono::hours>(elapsed);
if (hours != 0h)
{
formatted << hours.count() << " [h] ";
++usedUnits;
}
auto minutes = std::chrono::duration_cast<std::chrono::minutes>(elapsed % 1h);
if (minutes != 0min)
{
formatted << minutes.count() << " [min] ";
++usedUnits;
}
auto seconds = std::chrono::duration_cast<std::chrono::seconds>(elapsed % 1min);
if (seconds != 0min && usedUnits != maxUnits)
{
formatted << seconds.count() << " [s] ";
++usedUnits;
}
auto milliseconds = std::chrono::duration_cast<std::chrono::milliseconds>(elapsed % 1s);
if (milliseconds != 0ms && usedUnits != maxUnits)
{
formatted << milliseconds.count() << " [ms] ";
++usedUnits;
}
auto microseconds = std::chrono::duration_cast<std::chrono::microseconds>(elapsed % 1ms);
if (microseconds != 0us && usedUnits != maxUnits)
{
formatted << microseconds.count() << " [us] ";
++usedUnits;
}
auto nanoseconds = std::chrono::duration_cast<std::chrono::nanoseconds>(elapsed % 1us);
if (nanoseconds != 0us && usedUnits != maxUnits)
{
formatted << nanoseconds.count() << " [us] ";
++usedUnits;
}
return formatted.str();
}
int main() {
auto begin = std::chrono::system_clock::now();
// algorithm goes here
auto solution = "solution"; /* can be anything */
auto end = std::chrono::system_clock::now();
auto diff = end - begin;
using namespace std::chrono_literals;
std::cout << "Solution [" << solution << "] found in " << getTimeElapsed(1'234'567ns) << std::endl;
return 0;
}
如果您想更进一步,而不需要逃避类型系统,那么我建议您查阅Howard Hinnant's
template
库。该库是C ++ 20中新的date
功能的基础,并将字符串格式引入表中。只需以适合您的任何方式从库中包含chrono
,并按如下所示修改date.h
:std::string getTimeElapsed(std::chrono::nanoseconds elapsed, size_t maxUnits = 2)
{
using namespace std::chrono_literals;
std::ostringstream formatted("");
int usedUnits{};
auto hours = std::chrono::duration_cast<std::chrono::hours>(elapsed);
if (hours != 0h)
{
formatted << hours << " ";
++usedUnits;
}
auto minutes = std::chrono::duration_cast<std::chrono::minutes>(elapsed % 1h);
if (minutes != 0min)
{
formatted << minutes << " ";
++usedUnits;
}
auto seconds = std::chrono::duration_cast<std::chrono::seconds>(elapsed % 1min);
if (seconds != 0min && usedUnits != maxUnits)
{
formatted << seconds << " ";
++usedUnits;
}
auto milliseconds = std::chrono::duration_cast<std::chrono::milliseconds>(elapsed % 1s);
if (milliseconds != 0ms && usedUnits != maxUnits)
{
formatted << milliseconds << " ";
++usedUnits;
}
auto microseconds = std::chrono::duration_cast<std::chrono::microseconds>(elapsed % 1ms);
if (microseconds != 0us && usedUnits != maxUnits)
{
formatted << microseconds << " ";
++usedUnits;
}
auto nanoseconds = std::chrono::duration_cast<std::chrono::nanoseconds>(elapsed % 1us);
if (nanoseconds != 0us && usedUnits != maxUnits)
{
formatted << nanoseconds << " ";
++usedUnits;
}
return formatted.str();
}
使用与以前相同的值,结果将是:“ 1ms 234us”表示1'234'567ns。
关于c++ - 如何概括std::chrono::duration(s)?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/58352866/