我为大学班级编写了三种算法。
一个是蛮力，另一个是贪婪，最后一个是启发式。
我希望能够衡量每种算法需要花费多少时间。
我正在使用<chrono>库来实现这一目标
现在我的代码如下：
#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，堆积的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（？）
此功能看起来像这样：
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
答案 0 :(得分：1)
使用chrono时，最好的一般建议是仅在绝对必要时才逃避类型系统（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，这也意味着输出到流。
如果我们将自己保持在类型系统之内，我们会得到很多总是正确的转换。
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。呼叫getTimeElapsed尚不兼容chrono。
我并不完全满意，因此我们也支持duration中的getTimeElapsed：
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else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个{{ 1}}类型。
现在代码看起来好多了（仅保留相关更改）：
template太好了！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...
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");
}
}
}
}
...
成员，否则将导致编译器错误。
让我们限制一下getTimeElapsed：
.count()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，我认为（通过阅读评论）实际上是“我如何整理嵌套的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();
}
语句并仅打印出前两个非零单位”。
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，您需要编写一些代码以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，您仍然就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std::chrono::duration写成如下：
if将经过的总时间作为getTimeElapsed（您已经从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();
}
获得）并将其传递给std::chrono::nanoseconds。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end - begin为1'000'000'000ns，则结果为“ 1 [s]”;如果getTimeElapsed为1'234'568ns，则结果为“ 1 [ms] 234 [us]”。有尾随的空间，但我将留给您修复。
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：
elapsed如果您想更进一步，并且不需要逃避类型系统，那么我建议您浏览Howard Hinnant's date库。该库是C ++ 20中新的template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包括来自库的#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;
}
并按如下所示修改chrono：
date.h使用与以前相同的值，结果将是：“ 1ms 234us”持续1'234'567ns。