问题
I am wanting to print the running time of my functions. For some reason my timer always returns 0. Can anyone tell me why?
double RunningTime(clock_t time1, clock_t time2)
{
double t=time1 - time2;
double time = (t*1000)/CLOCKS_PER_SEC;
return time;
}
int main()
{
clock_t start_time = clock();
// some code.....
clock_t end_time = clock();
std::cout << "Time elapsed: " << double(RunningTime(end_time, start_time)) << " ms";
return 0;
}
I attempted to use gettimeofday
and it still returned 0.
double get_time()
{
struct timeval t;
gettimeofday(&t, NULL);
double d = t.tv_sec + (double) t.tv_usec/100000;
return d;
}
int main()
{
double time_start = get_time();
//Some code......
double time_end = get_time();
std::cout << time_end - time_start;
return 0;
}
Also tried using chrono
and it gave me all kinds of build errors:
- error: #error This file requires compiler and library support for the
upcoming ISO C++ standard, C++0x. This support is currently
experimental, and must be enabled with the -std=c++0x or -std=gnu++0x compiler options. - warning: 'auto' will change meaning in C++0x; please remove it
- error: ISO C++ forbids declaration of 't1' with no type error: 'std::chrono' has not been declared
error: request for member 'count' in '(t2 - t1)', which is of non-class type 'int'
int main() { auto t1 = std::chrono::high_resolution_clock::now();
//Some code...... auto t2 = std::chrono::high_resolution_clock::now(); std::cout << "Time elapsed: " << std::chrono::duration_cast<std::chrono::milliseconds>(t2-t1).count() << " milliseconds\n"; return 0; }
回答1:
A timer tick is approximately equal to 1/CLOCKS_PER_SEC second, which is a millisecond resolution. To see a real (non-zero) number, you should either invoke a very long-time function or use another library with a higher time resolution facility:
- new c++11x library
chrono
(use MSVS 2012) boost::chrono
(unfortunately, the library refers to a lot of others)- POSIX function gettimeofday, which gives you a 1 microsecond time resolution
回答2:
After lots of trial and error I went with gettimeofday
. Here is my code that I finally got to work properly.
double get_time()
{
struct timeval t;
gettimeofday(&t, NULL);
double d = t.tv_sec + (double) t.tv_usec/1000000;
return d;
}
int main()
{
double time_start = get_time();
//Some code.........
double time_end = get_time();
std::cout << time_end - time_start;
return 0;
}
回答3:
A solution I have been using lately uses C++11's lambda functionality to time any arbitrary function call or series of actions.
#include <ctime>
#include <iostream>
#include <functional>
void timeit(std::function<void()> func) {
std::clock_t start = std::clock();
func();
int ms = (std::clock() - start) / (double) (CLOCKS_PER_SEC / 1000);
std::cout << "Finished in " << ms << "ms" << std::endl;
}
int main() {
timeit([] {
for (int i = 0; i < 10; ++i) {
std::cout << "i = " << i << std::endl;
}
});
return 0;
}
来源:https://stackoverflow.com/questions/11111189/running-time-of-functions