I have following which stop execution of program after certain time.
#include <iostream>
#include<ctime>
using namespace std;
int main( )
{
time_t timer1;
time(&timer1);
time_t timer2;
double second;
while(1)
{
time(&timer2);
second = difftime(timer2,timer1);
//check if timediff is cross 3 seconds
if(second > 3)
{
return 0;
}
}
return 0;
}
Is above program would work if time increase from 23:59 to 00:01 ?
If there any other better way?
Provided you have C++11, you can have a look at this example:
#include <thread>
#include <chrono>
int main() {
std::this_thread::sleep_for (std::chrono::seconds(3));
return 0;
}
Alternatively I'd go with a threading library of your choice and use its Thread sleep function. In most cases it is better to send your thread to sleep instead of busy waiting.
time() returns the time since the Epoch (00:00:00 UTC, January 1, 1970), measured in seconds. Thus, the time of day does not matter.
You can use std::chrono::steady_clock in C++11. Check the example in the now static method for an example :
using namespace std::chrono;
steady_clock::time_point clock_begin = steady_clock::now();
std::cout << "printing out 1000 stars...\n";
for (int i=0; i<1000; ++i) std::cout << "*";
std::cout << std::endl;
steady_clock::time_point clock_end = steady_clock::now();
steady_clock::duration time_span = clock_end - clock_begin;
double nseconds = double(time_span.count()) * steady_clock::period::num / steady_clock::period::den;
std::cout << "It took me " << nseconds << " seconds.";
std::cout << std::endl;
Related
I have some program what must help me, but i cant handle timing.
Hire is a code:
#include <iostream>
using namespace std;
#include <time.h>
#include <Windows.h>
double diffclock(clock_t clock1) {
clock_t clock2 = clock();
double diffticks = clock1 - clock2;
double diffms = diffticks / (CLOCKS_PER_SEC / 1000);
return diffms;
}
int main()
{
int wait = 134;
clock_t fullbetween = clock();
for (int i = 0; i < 5; i++) {
Sleep(wait / 5);
cout << wait / 5 << " ";
}
cout << endl << "finish in " << diffclock(fullbetween) << " ms" << endl;
return 0;
}
C++ version. same result:
#include <iostream>
#include <chrono>
#include <ctime>
#include <thread>
#include <Windows.h>
int main()
{
int wait = 134;
auto start = std::chrono::system_clock::now();
for (int i = 0; i < 5; i++) {
std::this_thread::sleep_for(std::chrono::milliseconds(wait/5));
}
auto end = std::chrono::system_clock::now();
auto int_ms = std::chrono::duration_cast<std::chrono::milliseconds> (end - start);
std::cout << std::endl << "finish in " << int_ms.count() << " ms" << std::endl;
return 0;
}
134 / 5 = 26 is ok. But in last "cout" it shows that all that iteration taked about ~170ms, not 130 as expected. Why this is happening ?
Sry about my engl.
The documentation for the Sleep function at https://learn.microsoft.com/en-gb/windows/win32/api/synchapi/nf-synchapi-sleep says
Suspends the execution of the current thread until the time-out interval elapses.
The system clock "ticks" at a constant rate. If dwMilliseconds is less than the resolution of the system clock, the thread may sleep for less than the specified length of time. If dwMilliseconds is greater than one tick but less than two, the wait can be anywhere between one and two ticks, and so on.
Ticks are typically 15.6 ms on Windows systems (64 ticks per second), so 26 becomes 31.2.
This is the time after which it is possible for the suspended thread to become active again, there is no guarantee that it will start executing immediately. So your five sleeps become 156ms plus a little overhead.
The documentation continues with mitigations for this behaviour, and warnings that the mitigations will affect system power usage and so on.
To increase the accuracy of the sleep interval, call the timeGetDevCaps function to determine the supported minimum timer resolution and the timeBeginPeriod function to set the timer resolution to its minimum.
In std::this_thread::sleep_for documentation (found here)
It is stated that the function blocks the execution of the current
thread for at least the specified sleep_duration.
It may block for
longer than sleep_duration due to scheduling or resource contention
delays.
So your code will take at least 135ms to execute.
#include <iostream>
#include <chrono>
#include <time.h>
#include <stdio.h>
using namespace std;
using namesapce chrono;
int main() {
int f;
time_t start, end;
time (&start);
cin >> f;
time (&end);
double dif = difftime (end, start);
printf ("Elapsed time is %.2lf seconds.", dif );
}
Hello everyone, I am currently working on a C++ assignment and I am essentially required to have the user input something within 10 seconds. I managed to find out how to count the time by the second but I need it to be by milliseconds because I have to find out how many milliseconds above 10 seconds was elapsed. I am not that experienced with C++ and would very much appreciate any suggestions that may help steer me in the right direction. Thanks a lot
Something along these lines...
#include <iostream>
#include <chrono>
auto start(std::chrono::high_resolution_clock::now());
// Code...
auto end(std::chrono::high_resolution_clock::now());
auto duration(std::chrono::duration_cast<std::chrono::milliseconds>(end - start));
std::cout << "Duration: " << duration.count() << " ms\n";
in C++11 and more recent Standard revisions:
#include <chrono>
using namespace std::chrono;
auto start = high_resolution_clock::now();
// something to measure
auto end = high_resolution_clock::now();
duration<double> diff = end - start; // this is in ticks
milliseconds d = duration_cast<milliseconds>(diff); // ticks to time
std::cout << diff.count() << "s\n";
std::cout << d.count() << "ms\n";
in previous to that :
<sys/time.h>
struct timeval tp;
gettimeofday(&tp, NULL);
long int ms = tp.tv_sec * 1000 + tp.tv_usec / 1000;
Also you can use this simple snippet code to benchmark your code blocks :
using namespace std::chrono;
class benchmark {
public:
time_point<high_resolution_clock> t0, t1;
unsigned int *d;
benchmark(unsigned int *res) : d(res) {
t0 = high_resolution_clock::now();
}
~benchmark() { t1 = high_resolution_clock::now();
milliseconds dur = duration_cast<milliseconds>(t1 - t0);
*d = dur.count();
}
};
// one way to use it can be :
#define BENCH(TITLE,CODEBLOCK) \
unsigned int __time__##__LINE__ = 0; \
{ benchmark bench(&__time__##__LINE__); \
CODEBLOCK \
} \
printf("%s took %dms\n",(TITLE),__time__##__LINE__);
int main(void) {
BENCH("TITLE",{
for(int n = 0; n < testcount; n++ )
int a = n % 3;
});
return 0;
}
The following complete program shows how this can be done, by using the std::chrono facilities added in C++11:
#include <iostream>
#include <chrono>
#include <thread>
int main(int argc, char *argv[]) {
// Check args.
if (argc != 2) {
std::cerr << "Usage: testprog <sleepTime>" << std::endl;
return 1;
}
// Create a millisecond sleep time from argument.
auto sleepTime = strtoul(argv[1], nullptr, 10);
sleepTime = sleepTime * 1234 + 1000;
std::cout << "Given '" << argv[1] <<
"', should sleep for about " << sleepTime <<
"ms ... " << std::flush;
// Get the start time, then wait for a bit.
auto startTime(std::chrono::high_resolution_clock::now());
std::this_thread::sleep_for(std::chrono::milliseconds(sleepTime));
// Get end time, work out and print duration.
auto endTime(std::chrono::high_resolution_clock::now());
auto duration(std::chrono::duration_cast<std::chrono::milliseconds>
(endTime - startTime));
std::cout << "that took " << duration.count() << "ms." << std::endl;
}
Running this with the following bash test command:
for i in {0..10} ; do ./testprog $i ; done
gives you the results you would expect:
Given '0', should sleep for about 1000ms ... that took 1000ms.
Given '1', should sleep for about 2234ms ... that took 2235ms.
Given '2', should sleep for about 3468ms ... that took 3469ms.
Given '3', should sleep for about 4702ms ... that took 4703ms.
Given '4', should sleep for about 5936ms ... that took 5937ms.
Given '5', should sleep for about 7170ms ... that took 7170ms.
Given '6', should sleep for about 8404ms ... that took 8404ms.
Given '7', should sleep for about 9638ms ... that took 9638ms.
Given '8', should sleep for about 10872ms ... that took 10872ms.
Given '9', should sleep for about 12106ms ... that took 12106ms.
Given '10', should sleep for about 13340ms ... that took 13340ms.
The important lines in that code are really just the ones that get the start and end time-points then work out the duration between them. They can be boiled down to:
#include <chrono>
auto startTime(std::chrono::high_resolution_clock::now());
// Do whatever you want to time.
auto endTime(std::chrono::high_resolution_clock::now());
auto duration(std::chrono::duration_cast<std::chrono::milliseconds>
(endTime - startTime));
auto elapsedMs = duration.count();
While I realize this is probably one of many identical questions, I can't seem to figure out how to properly use std::chrono. This is the solution I cobbled together.
#include <stdlib.h>
#include <iostream>
#include <chrono>
typedef std::chrono::high_resolution_clock Time;
typedef std::chrono::milliseconds ms;
float startTime;
float getCurrentTime();
int main () {
startTime = getCurrentTime();
std::cout << "Start Time: " << startTime << "\n";
while(true) {
std::cout << getCurrentTime() - startTime << "\n";
}
return EXIT_SUCCESS;
}
float getCurrentTime() {
auto now = Time::now();
return std::chrono::duration_cast<ms>(now.time_since_epoch()).count() / 1000;
}
For some reason, this only ever returns integer values as the difference, which increments upwards at rate of 1 per second, but starting from an arbitrary, often negative, value.
What am I doing wrong? Is there a better way of doing this?
Don't escape the chrono type system until you absolutely have to. That means don't use .count() except for I/O or interacting with legacy API.
This translates to: Don't use float as time_point.
Don't bother with high_resolution_clock. This is always a typedef to either system_clock or steady_clock. For more portable code, choose one of the latter.
.
#include <iostream>
#include <chrono>
using Time = std::chrono::steady_clock;
using ms = std::chrono::milliseconds;
To start, you're going to need a duration with a representation of float and the units of seconds. This is how you do that:
using float_sec = std::chrono::duration<float>;
Next you need a time_point which uses Time as the clock, and float_sec as its duration:
using float_time_point = std::chrono::time_point<Time, float_sec>;
Now your getCurrentTime() can just return Time::now(). No fuss, no muss:
float_time_point
getCurrentTime() {
return Time::now();
}
Your main, because it has to do the I/O, is responsible for unpacking the chrono types into scalars so that it can print them:
int main () {
auto startTime = getCurrentTime();
std::cout << "Start Time: " << startTime.time_since_epoch().count() << "\n";
while(true) {
std::cout << (getCurrentTime() - startTime).count() << "\n";
}
}
This program does a similar thing. Hopefully it shows some of the capabilities (and methodology) of std::chrono:
#include <iostream>
#include <chrono>
#include <thread>
int main()
{
using namespace std::literals;
namespace chrono = std::chrono;
using clock_type = chrono::high_resolution_clock;
auto start = clock_type::now();
for(;;) {
auto first = clock_type::now();
// note use of literal - this is c++14
std::this_thread::sleep_for(500ms);
// c++11 would be this:
// std::this_thread::sleep_for(chrono::milliseconds(500));
auto last = clock_type::now();
auto interval = last - first;
auto total = last - start;
// integer cast
std::cout << "we just slept for " << chrono::duration_cast<chrono::milliseconds>(interval).count() << "ms\n";
// another integer cast
std::cout << "also known as " << chrono::duration_cast<chrono::nanoseconds>(interval).count() << "ns\n";
// floating point cast
using seconds_fp = chrono::duration<double, chrono::seconds::period>;
std::cout << "which is " << chrono::duration_cast<seconds_fp>(interval).count() << " seconds\n";
std::cout << " total time wasted: " << chrono::duration_cast<chrono::milliseconds>(total).count() << "ms\n";
std::cout << " in seconds: " << chrono::duration_cast<seconds_fp>(total).count() << "s\n";
std::cout << std::endl;
}
return 0;
}
example output:
we just slept for 503ms
also known as 503144616ns
which is 0.503145 seconds
total time wasted: 503ms
in seconds: 0.503145s
we just slept for 500ms
also known as 500799185ns
which is 0.500799 seconds
total time wasted: 1004ms
in seconds: 1.00405s
we just slept for 505ms
also known as 505114589ns
which is 0.505115 seconds
total time wasted: 1509ms
in seconds: 1.50923s
we just slept for 502ms
also known as 502478275ns
which is 0.502478 seconds
total time wasted: 2011ms
in seconds: 2.01183s
I have a starting timepoint in milliseconds like so:
using namespace std::chrono;
typedef time_point<system_clock, milliseconds> MyTimePoint;
MyTimePoint startTimePoint = time_point_cast<MyTimePoint::duration>(system_clock::time_point(steady_clock::now()));
Now I will have a certain number of hours that I want to add or subtract to the startTimePoint.
int numHours = -5//or 5 etc (Can be a plus or minus number)
How can I add this abount of time to the original startTimePoint??
If you want to add five hours to startTimePoint, it's boringly simple:
startTimePoint += hours(5); // from the alias std::chrono::hours
Live example.
By the way, you're trying to convert a steady_clock::now() into a system_clock::time_point, which shouldn't even compile. Change the steady_clock::now() to system_clock::now() and you should be good to go.
Here I have used time in minutes you can go for anything that you want from the user.
So the below is the simple programme using chrono
#include <iostream>
#include <chrono>
using namespace std;
int main() {
using clock = std::chrono::system_clock;
clock::time_point nowp = clock::now();
cout<<"Enter the time that you want to add in minutes"<<endl;
int time_min;
cin>>time_min;
cin.ignore();
clock::time_point end = nowp + std::chrono::minutes(time_min);
time_t nowt = clock::to_time_t ( nowp );
time_t endt = clock::to_time_t ( end);
std::cout << " " << ctime(&nowt) << "\n";
std::cout << ctime(&endt) << std::endl;
return 0;
}
Convert time_point to duration or duration to time_point without intermediate.
It is inherently impossible to convert a time_point to duration or back directly.
Many examples use time_t as intermediate, which is a fine method.
I use the method that uses the time_point 'zero' as a helper.
#include <iostream>
#include <chrono>
#include <thread>
using namespace std;
int main(int argc, char *argv[])
{
using namespace std::chrono;
system_clock::time_point zero; // initialised to zero in constructor
system_clock::time_point tp_now; // now as time_point
duration<int, ratio<1>> dur_now; // now as duration
system_clock::time_point tp_future; // calculated future as time_point
// The objective is to sleep_until the system time is at the next 5 minutes
// boundary (e.g. time is 09:35)
tp_now = system_clock::now(); // What time is it now?
cout << "tp_now = " << tp_now.time_since_epoch().count() << endl;
// It is not possible to assign a time_point directly to a duration.
// but the difference between two time_points can be cast to duration
dur_now = duration_cast<seconds>(tp_now-zero); // subtract nothing from time_point
cout << "dur_now = " << dur_now.count() << endl;
// Instead of using seconds granularity, I want to use 5 minutes
// so I define a suitable type: 5 minutes in seconds
typedef duration<int,ratio<5*60>> dur5min;
// When assigning the time_point (ok: duration) is truncated to the nearest 5min
dur5min min5 = duration_cast<dur5min>(tp_now-zero); // (Yes, I do it from time_point again)
cout << "min5 ('now' in 5min units) = " << min5.count() << endl;
// The next 5 min time point is
min5 += dur5min{1};
cout << "min5 += dur5min{1} = " << min5.count() << endl;
// It is not possible to assign a duration directly to a time_point.
// but I can add a duration to a time_point directly
tp_future = zero + min5;
cout << "tp_future = " << tp_future.time_since_epoch().count() << endl;
// to be used in e.g. sleep_until
// std::this_thread::sleep_until(tp_future);
return 0;
}
Thanks to Carsten's solution I managed to create function:
#include <chrono>
auto getTimeDurationMovedWith(std::chrono::hours hours2move)
{
using namespace std::chrono;
auto current_time = system_clock::now();
decltype(current_time) zeroTime; // no better solution to move time found in stackoverflow
return chrono::duration_cast<microseconds>(
current_time - zeroTime + hours(hours2move));
}
And it can be used like that:
auto tmp = getTimeDurationMovedWith(chrono::hours(-10));
cout << tmp.count() << endl;
How do I call clock() in C++?
For example, I want to test how much time a linear search takes to find a given element in an array.
#include <iostream>
#include <cstdio>
#include <ctime>
int main() {
std::clock_t start;
double duration;
start = std::clock();
/* Your algorithm here */
duration = ( std::clock() - start ) / (double) CLOCKS_PER_SEC;
std::cout<<"printf: "<< duration <<'\n';
}
An alternative solution, which is portable and with higher precision, available since C++11, is to use std::chrono.
Here is an example:
#include <iostream>
#include <chrono>
typedef std::chrono::high_resolution_clock Clock;
int main()
{
auto t1 = Clock::now();
auto t2 = Clock::now();
std::cout << "Delta t2-t1: "
<< std::chrono::duration_cast<std::chrono::nanoseconds>(t2 - t1).count()
<< " nanoseconds" << std::endl;
}
Running this on ideone.com gave me:
Delta t2-t1: 282 nanoseconds
clock() returns the number of clock ticks since your program started. There is a related constant, CLOCKS_PER_SEC, which tells you how many clock ticks occur in one second. Thus, you can test any operation like this:
clock_t startTime = clock();
doSomeOperation();
clock_t endTime = clock();
clock_t clockTicksTaken = endTime - startTime;
double timeInSeconds = clockTicksTaken / (double) CLOCKS_PER_SEC;
On Windows at least, the only practically accurate measurement mechanism is QueryPerformanceCounter (QPC). std::chrono is implemented using it (since VS2015, if you use that), but it is not accurate to the same degree as using QueryPerformanceCounter directly. In particular it's claim to report at 1 nanosecond granularity is absolutely not correct. So, if you're measuring something that takes a very short amount of time (and your case might just be such a case), then you should use QPC, or the equivalent for your OS. I came up against this when measuring cache latencies, and I jotted down some notes that you might find useful, here;
https://github.com/jarlostensen/notesandcomments/blob/master/stdchronovsqcp.md
#include <iostream>
#include <ctime>
#include <cstdlib> //_sleep() --- just a function that waits a certain amount of milliseconds
using namespace std;
int main()
{
clock_t cl; //initializing a clock type
cl = clock(); //starting time of clock
_sleep(5167); //insert code here
cl = clock() - cl; //end point of clock
_sleep(1000); //testing to see if it actually stops at the end point
cout << cl/(double)CLOCKS_PER_SEC << endl; //prints the determined ticks per second (seconds passed)
return 0;
}
//outputs "5.17"
You can measure how long your program works. The following functions help measure the CPU time since the start of the program:
C++ (double)clock() / CLOCKS_PER_SEC with ctime included.
Python time.clock() returns floating-point value in seconds.
Java System.nanoTime() returns long value in nanoseconds.
My reference: algorithms toolbox week 1 course part of data structures and algorithms specialization by University of California San Diego & National Research University Higher School of Economics
So you can add this line of code after your algorithm:
cout << (double)clock() / CLOCKS_PER_SEC;
Expected Output: the output representing the number of clock ticks per second
Probably you might be interested in timer like this :
H : M : S . Msec.
the code in Linux OS:
#include <iostream>
#include <unistd.h>
using namespace std;
void newline();
int main() {
int msec = 0;
int sec = 0;
int min = 0;
int hr = 0;
//cout << "Press any key to start:";
//char start = _gtech();
for (;;)
{
newline();
if(msec == 1000)
{
++sec;
msec = 0;
}
if(sec == 60)
{
++min;
sec = 0;
}
if(min == 60)
{
++hr;
min = 0;
}
cout << hr << " : " << min << " : " << sec << " . " << msec << endl;
++msec;
usleep(100000);
}
return 0;
}
void newline()
{
cout << "\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n";
}