I am trying to find the average UTC time of when a function was called. So I do:
boost::posix_time::ptime current_time_before(boost::posix_time::microsec_clock::universal_time());
DoStuff();
boost::posix_time::ptime current_time_after(boost::posix_time::microsec_clock::universal_time());
How do I go about calculating the averages between these two times?
I tried:
double time_avg = (current_time_before+current_time_after)*0.5;
But I get an error on a linux system that seems to have a problem with "+" but not "-" .
Thank you for your help.
Just... write it naturally?
ptime midpoint(ptime const& a, ptime const& b) {
return a + (b-a)/2; // TODO check for special case `b==a`
}
Live demo:
Live On Coliru
#include <boost/date_time/posix_time/posix_time.hpp>
using boost::posix_time::ptime;
ptime midpoint(ptime const& a, ptime const& b) {
return a + (b-a)/2;
}
int main() {
ptime a = boost::posix_time::second_clock::local_time();
ptime b = a + boost::posix_time::hours(3);
std::cout << "Mid of " << a << " and " << b << " is " << midpoint(a,b) << "\n";
std::swap(a,b);
std::cout << "Mid of " << a << " and " << b << " is " << midpoint(a,b) << "\n";
}
Prints
Mid of 2016-Sep-15 11:17:10 and 2016-Sep-15 14:17:10 is 2016-Sep-15 12:47:10
Mid of 2016-Sep-15 14:17:10 and 2016-Sep-15 11:17:10 is 2016-Sep-15 12:47:10
Related
I'm trying to write a simple single header benchmarker and I understand that std::clock will give me the time that a process (thread) is in actual use.
So, given the following simplified program:
nt main() {
using namespace std::literals::chrono_literals;
auto start_cpu = std::clock();
auto start_wall = std::chrono::high_resolution_clock::now();
// clobber();
std::this_thread::sleep_for(1s);
// clobber();
auto finish_cpu = std::clock();
auto finish_wall = std::chrono::high_resolution_clock::now();
std::cerr << "cpu: "
<< start_cpu << " " << finish_cpu << " "
<< (finish_cpu - start_cpu) / (double)CLOCKS_PER_SEC << " s" << std::endl;
std::cerr << "wall: "
// << FormatTime(start_wall) << " " << FormatTime(finish_wall) << " "
<< (finish_wall - start_wall) / 1.0s << " s" << std::endl;
return 0;
}
Demo
We get the following output:
cpu: 4820 4839 1.9e-05 s
wall: 1.00007 s
I just want to clarify that the cpu time is the time that it executes the code that is not actually the sleep_for code as that is actually done by the kernel which std::clock doesn't track. So to confirm, I changed what I was timing:
int main() {
using namespace std::literals::chrono_literals;
int value = 0;
auto start_cpu = std::clock();
auto start_wall = std::chrono::high_resolution_clock::now();
// clobber();
for (int i = 0; i < 1000000; ++i) {
srand(value);
value = rand();
}
// clobber();
std::cout << "value = " << value << std::endl;
auto finish_cpu = std::clock();
auto finish_wall = std::chrono::high_resolution_clock::now();
std::cerr << "cpu: "
<< start_cpu << " " << finish_cpu << " "
<< (finish_cpu - start_cpu) / (double)CLOCKS_PER_SEC << " s" << std::endl;
std::cerr << "wall: "
// << FormatTime(start_wall) << " " << FormatTime(finish_wall) << " "
<< (finish_wall - start_wall) / 1.0s << " s" << std::endl;
return 0;
}
Demo
This gave me an output of:
cpu: 4949 1398224 1.39328 s
wall: 2.39141 s
value = 354531795
So far, so good. I then tried this on my windows box running MSYS2's g++ compiler. The output for the last program gave me:
value = 0
cpu: 15 15 0 s
wall: 0.0080039 s
std::clock() is always outputting 15? Is the compiler implementation of std::clock() broken?
Seems that I assumed that CLOCKS_PER_SEC would be the same. However, on the MSYS2 compiler, it was 1000x less then on godbolt.org.
I am trying to make a text game where there is a timer and once the game was finished before or in 60 seconds, there is a bonus points. However, I have no idea how can I get the value or the time from using the chrono without cout-ing it. I want to use the value for calculating the bonus point. i can cout the value through the .count() but I cannot get that value to use for the condition part.
here's my code for the scoring part:
void Game::score(auto start, auto end) {
int bonus = 0;
int total = 0;
string name;
box();
gotoxy(10,8); cout << "C O N G R A T U L A T I O N S";
gotoxy(15,10); cout << "You have successfully accomplished all the levels!";
gotoxy(15,11); cout << "You are now a certified C-O-N-N-E-C-T-o-r-I-s-T" << char(002) << char(001);
gotoxy(20,13); cout << "= = = = = = = = = = GAME STATS = = = = = = = = = =";
gotoxy(25,15); cout << "Time Taken: " << chrono::duration_cast<chrono::seconds>(end - start).count() << " seconds";
gotoxy(25,16); cout << "Points: " << pts << " points";
if (chrono::duration_cast<chrono::seconds>(end - start).count() <= 60) {
bonus == 5000;
} else if (chrono::duration_cast<chrono::seconds>(end - start).count() <= 90) {
bonus == 3000;
} else if (chrono::duration_cast<chrono::seconds>(end - start).count() <= 120) {
bonus == 1000;
}
gotoxy(30,17); cout << "Bonus Points (Time Elapsed): " << bonus;
total = pts + bonus;
gotoxy(25,18); cout << "Total Points: " << total << " points";
gotoxy(20,20); cout << "Enter your name: ";
cin >> name;
scoreB.open("scoreboard.txt",ios::app);
scoreB << name << "\t" << total << "\n";
scoreB.close();
}
You should really use the chrono literals for comparing durations. See example here:
#include <chrono>
#include <iostream>
#include <thread>
using Clock = std::chrono::system_clock;
void compareTimes(std::chrono::time_point<Clock> startTime,
std::chrono::time_point<Clock> finishTime) {
using namespace std::chrono_literals;
std::chrono::duration<float> elapsed = finishTime - startTime;
std::cout << "elapsed = " << elapsed.count() << "\n";
if (elapsed > 10ms) {
std::cout << "over 10ms\n";
}
if (elapsed < 60s) {
std::cout << "under 60s\n";
}
}
int main() {
using namespace std::chrono_literals;
auto startTime = Clock::now();
std::this_thread::sleep_for(20ms);
auto finishTime = Clock::now();
compareTimes(startTime, finishTime);
return 0;
}
Demo: https://godbolt.org/z/hqv58acoY
I'd like to add a timestamp to certain outputs to the std::cout / std::cerr ostreams, without using modified standard streams, like so:
std::cerr << timestamp << "Warning!\n";
or so:
std::cerr << timestamp() << "Warning!\n";
The output should look like this:
[2020-01-23 17:40:15 CET] Warning!
But I'm really not happy with what I've come up with:
class TimeStamp {};
std::ostream &operator<<(std::ostream &stream, const TimeStamp &ts)
{
std::time_t t = std::time(nullptr);
stream << "[" << std::put_time(std::localtime(&t), "%F %T %Z") << "] ";
return stream;
}
TimeStamp ts;
int main()
{
std::cerr << ts << "Warning!\n";
std::cerr << ts << "Another warning!\n";
}
So I'm basically defining an empty class, using a global declaration and overloading the '<<' operator. This feels wrong. A static function like timestamp() is probably better suited, but I'm not quite sure how to go on about this. All the examples I've found online used the overloaded '<<' operator, but it usually made more sense to do so, because some class state was output. Can I locally create an ostream and return that in the function?
There's nothing wrong with the way you've done it. But if you're looking for alternatives, you could create an ostream wrapper:
class Logger {
private:
std::ostream &stream;
void print_time() {
std::time_t t = std::time(nullptr);
stream << "[" << std::put_time(std::localtime(&t), "%F %T %Z") << "] ";
}
public:
//Maybe also take options for how to log?
Logger(std::ostream &stream) : stream(stream) { }
template <typename T>
std::ostream &operator<<(const T &thing) {
print_time();
return stream << thing;
}
};
int main()
{
Logger log(std::cerr);
log << "Warning!" << std::endl;
log << "Another warning!" << std::endl;
}
See it run here: https://ideone.com/YRawuQ
If you're just looking for a standalone function which is what I understood from a "static function like timestamp()" you can just return the date as a string:
std::string timeStamp(){
std::ostringstream strStream;
std::time_t t = std::time(nullptr);
strStream<< "[" << std::put_time(std::localtime(&t), "%F %T %Z") << "] ";
return strStream.str();
}
int main(){
std::cout<<timeStamp()<<" Testing!";
return 0;
}
Remember to include sstream
You can use standard std::chrono::time_point class from the date and time library to represent the timestamp. Then, you need to convert that timestamp to std::time_t type and, eventually, convert the date and time information from a given calendar time to a character string according to the format string.
auto const now = std::chrono::system_clock::now();
auto now_time = std::chrono::system_clock::to_time_t(now);
std::cout << std::put_time(std::localtime(&now_time), "%F %T") << std::endl;
For those who want to know more...
You can use the source_location class that represents certain information about the source code, such as file names, line numbers, and function names. It is being merged into ISO C++ and is available for use.
Full code
#include <ctime>
#include <chrono>
#include <iomanip>
#include <iostream>
#include <string_view>
#include <experimental/source_location>
void error(std::string_view const& message,
std::ostream& os = std::cout,
std::experimental::source_location const& location = std::experimental::source_location::current()) {
auto const now = std::chrono::system_clock::now();
auto now_time = std::chrono::system_clock::to_time_t(now);
os << "[" << std::put_time(std::localtime(&now_time), "%F %T") << "] "
<< "[INFO] "
<< location.file_name() << ":"
<< location.line() << " "
<< message << '\n';
}
void info(std::string_view const& message,
std::ostream& os = std::cout,
std::experimental::source_location const& location = std::experimental::source_location::current()) {
auto const now = std::chrono::system_clock::now();
auto now_time = std::chrono::system_clock::to_time_t(now);
os << "[" << std::put_time(std::localtime(&now_time), "%F %T") << "] "
<< "[INFO] "
<< location.file_name() << ":"
<< location.line() << " "
<< message << '\n';
}
int main() {
error("Some error");
info("Some info");
// or
error("Some error 2", std::cerr);
info("Some info 2", std::cerr);
return 0;
}
Check it out live
I have been searching on Google an in this forum for a while, but I could not find any answer or tip for my problem. Tutorials couldn't help me either...
I want to redistribute some points, stored in a vector p_org. (x-value is stored as double).
Therefore I have the function distribute, which is defined in maths.h
distribute_tanh(&p_org_temp,&p_new_temp,iz,spacing[0],spacing[1],l_rot[(kk+1)*iz-2],status);
The function distribute_tanh does look like this:
inline void distribute_tanh (std::vector<double> *p_org, std::vector<double> *p_new, const int n_points, double spacing_begin, double spacing_end, const double total_length, double status){
//if status == 0: FLAP, if status == 1: SLAT
std::cout << "spacing_begin: " << spacing_begin << " spacing_end: " << spacing_end << std::endl;
double s_begin = spacing_begin / total_length;
double s_end = spacing_end / total_length;
double A = sqrt(s_end/s_begin);
double B = 1 / (sqrt(s_end*s_begin)*n_points);
std::cout << "A: " << A << " B: " << B << std::endl;
std::vector<double> u (n_points);
std::vector<double> sn (n_points);
double dx;
double dy;
std::cout << "Control at the beginning: p_org: " << (p_org) << " p_new: " << (p_new) << " n_points: " << n_points << " s_begin: " << s_begin << " s_end: " << s_end << " total_length: " << total_length << std::endl;
//problem no. 1
for (int i=0;i<n_points;i++){
if (B > 1.001) {
if (B < 2.7829681) {
double Bq=B-1;
dy=sqrt(6*Bq)*(1-0.15*Bq+0.057321429*pow(Bq,2)-0.024907295*pow(Bq,3)+0.0077424461*pow(Bq,4)-0.0010794123*pow(Bq,5));
} else if (B > 2.7829681) {
double Bv=log(B);
double Bw=1/B-0.028527431;
dy=Bv+(1+1/Bv)*log(2*Bv)-0.02041793+0.24902722*Bw+1.9496443*pow(Bw,2)-2.6294547*pow(Bw,3)+8.56795911*pow(Bw,4);
}
u[i]=0.5+(tanh(dy*(i*(1.0/n_points)-0.5))/(2*tanh(dy/2)));
}
else if (B < 0.999) {
if (B < 0.26938972) {
dx=M_PI*(1-B+pow(B,2)-(1+(pow(M_PI,2))/6)*pow(B,3)+6.794732*pow(B,4)-13.205501*pow(B,5)+11.726095*pow(B,6));
} else if (B > 0.26938972) {
double Bq=1-B;
dx=sqrt(6*Bq)*(1+0.15*Bq+0.057321429*pow(Bq,2)+0.048774238*pow(Bq,3)-0.053337753*pow(Bq,4)+0.075845134*pow(Bq,5));
}
u[i]=0.5+(tan(dx*(i*(1.0/n_points)-0.5))/(2*tan(dx/2)));
}
else {
u[i]=i*(1.0/n_points)*(1+2*(B-1)*(i*(1.0/n_points)-0.5)*(1-i*(1.0/n_points)));
}
sn[i]=u[i]/(A+(1.0-A)*u[i]);
std::cout << "sn(i): " << sn[i] << std::endl;
std::cout << "p_org[n_points]: " << &p_org[n_points-1] << std::endl;
if(status==0){
//p_new[i]=p_org[0]+(total_length*sn[i]);
std::cout << "FLAP maths.h" << std::endl;
}
//Here is the problem no. 2
else if(status==1){
//p_new[i]=p_org[0]-(total_length*sn[i]);
std::cout << "SLAT maths.h" << std::endl;
}
//std::cout << "p_new in math: " << p_new << std::endl;
}
}
My problem is, that I am unable to access the value of p_org or p_new. At the beginning I would like to give out the value of p_org and p_new. If I try it with a *, the compiler is complaining: error: no operator "<<" matches these operands
operand types are: std::basic_ostream> << std::vector>
std::cout << "Control at the beginning: p_org: " << (*p_org) << " p_new: " << (*p_new) << " n_points: " << n_points << " s_begin: " << s_begin << " s_end: " << s_end << " total_length: " << total_length << std::endl;
If I leave the * off, I get the addresses of p_org and p_new.
At the end of the code I would like to write the new value to p_new. If I use * to access the value, the compiler is complaining, if I leave it off, its complaining too with the following message:
error: no operator "-" matches these operands
operand types are: std::vector<double, std::allocator<double>> - double
p_new[i]=p_org[0]-(total_length*sn[i]);
^
I tried to understand both problems, but until now I had no success.
Thanks for your advice.
Your issue with the compiler error can be cut down to a very simple program.
#include <vector>
void foo(std::vector<int>* pV)
{
pV[0] = 10; // error.
}
int main()
{
std::vector<int> v(10);
foo(&v);
}
The issue is that operator[] as done above works for objects and references, not pointers. Since pv is a pointer, you must dereference it first to obtain the object, and then apply [] to the dereferenced pointer.
void foo(std::vector<int>* pV)
{
(*pV)[0] = 10; // No error
}
The other form of calling operator[] can be also used, but is a bit more verbose:
void foo(std::vector<int>* pV)
{
pv->operator[](0) = 10; // No error
}
However, to alleviate having to do this, pass the vector by reference. Then the "normal" way of using operator[] can be used.
#include <vector>
void foo(std::vector<int>& pV)
{
pV[0] = 10; // No error.
}
int main()
{
std::vector<int> v(10);
foo(v);
}
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
class TimeUnit
{
public:
TimeUnit(int m, int s)
{
this -> minutes = m;
this -> seconds = s;
}
string ToString()
{
ostringstream o;
o << minutes << " minutes and " << seconds << " seconds." << endl;
return o.str();
}
void Simplify()
{
if (seconds >= 60)
{
minutes += seconds / 60;
seconds %= 60;
}
}
TimeUnit Add(TimeUnit t2)
{
TimeUnit t3;
t3.seconds = seconds + t2.seconds;
if(t3.seconds >= 60)
{
t2.minutes += 1;
t3.seconds -= 60;
}
t3.minutes = minutes + t2.minutes;
return t3;
}
private:
int minutes;
int seconds;
};
int main(){
cout << "Hello World!" << endl;
TimeUnit t1(2,30);
cout << "Time1:" << t1.ToString() << endl;
TimeUnit t2(3,119);
cout << "Time2:" << t2.ToString();
t2.Simplify();
cout << " simplified: " << t2.ToString() << endl;
cout << "Added: " << t1.Add(t2).ToString() << endl;
//cout << " t1 + t2: " << (t1 + t2).ToString() << endl;
/*cout << "Postfix increment: " << (t2++).ToString() << endl;
cout << "After Postfix increment: " << t2.ToString() << endl;
++t2;
cout << "Prefix increment: " << t2.ToString() << endl;*/
}
I'm having problems with my Add method. Xcode is giving me this error: "No matching constructor for initialization of TimeUnit"
Could someone please tell me what I am doing wrong? I've literally tried everything that I know how to do, but I can't even get it to compile with this method.
Here are the instructions from my professor:
The TimeUnit class should be able to hold a time consisting of Minutes
and Seconds. It should have the following methods:
A constructor that takes a Minute and Second as parameters ToString()
- Should return the string equivilant of the time. "M minutes S seconds." Test1 Simplify() - This method should take the time and
simplify it. If the seconds is 60 seconds or over, it should reduce
the seconds down to below 60 and increase the minutes. For example, 2
Min 121 seconds should become 4 minutes 1 second. Test2 Add(t2) -
Should return a new time that is the simplified addition of the two
times Test3 operator + should do the same thing as Add Test4 pre and
postfix ++: should increase the time by 1 second and simplify Test5
In your TimeUnit::Add function, you tried to initialize t3 with default constructor. However, your TimeUnit doesn't have one:
TimeUnit Add(TimeUnit t2)
{
TimeUnit t3; ///<<<---- here
///.....
}
Try update TimeUnit::Add to this way:
TimeUnit Add(const TimeUnit& t2)
{
return TimeUnit(this->minutes+t2.minutes, this->seconds+t2.seconds);
}
The specific problem is because there is no TimeUnit::TimeUnit() defined, only TimeUnit(const int &m, const int &s).