Execution time of C program - c

Question

I have a C program that aims to be run in parallel on several processors. I need to be able to record the execution time (which could be anywhere from 1 second to several minutes). I have searched for answers, but they all seem to suggest using the clock() function, which then involves calculating the number of clocks the program took divided by the Clocks_per_second value.
I'm not sure how the Clocks_per_second value is calculated?
In Java, I just take the current time in milliseconds before and after execution.
Is there a similar thing in C? I've had a look, but I can't seem to find a way of getting anything better than a second resolution.
I'm also aware a profiler would be an option, but am looking to implement a timer myself.
Thanks

Answer 1

CLOCKS_PER_SEC is a constant which is declared in <time.h>. To get the CPU time used by a task within a C application, use:
clock_t begin = clock();
/* here, do your time-consuming job */
clock_t end = clock();
double time_spent = (double)(end - begin) / CLOCKS_PER_SEC;
Note that this returns the time as a floating point type. This can be more precise than a second (e.g. you measure 4.52 seconds). Precision depends on the architecture; on modern systems you easily get 10ms or lower, but on older Windows machines (from the Win98 era) it was closer to 60ms.
clock() is standard C; it works "everywhere". There are system-specific functions, such as getrusage() on Unix-like systems.
Java's System.currentTimeMillis() does not measure the same thing. It is a "wall clock": it can help you measure how much time it took for the program to execute, but it does not tell you how much CPU time was used. On a multitasking systems (i.e. all of them), these can be widely different.

Answer 2

If you are using the Unix shell for running, you can use the time command.
doing
$ time ./a.out
assuming a.out as the executable will give u the time taken to run this

Answer 3

In plain vanilla C:
#include <time.h>
#include <stdio.h>
int main()
{
clock_t tic = clock();
my_expensive_function_which_can_spawn_threads();
clock_t toc = clock();
printf("Elapsed: %f seconds\n", (double)(toc - tic) / CLOCKS_PER_SEC);
return 0;
}

Answer 4

You functionally want this:
#include <sys/time.h>
struct timeval tv1, tv2;
gettimeofday(&tv1, NULL);
/* stuff to do! */
gettimeofday(&tv2, NULL);
printf ("Total time = %f seconds\n",
(double) (tv2.tv_usec - tv1.tv_usec) / 1000000 +
(double) (tv2.tv_sec - tv1.tv_sec));
Note that this measures in microseconds, not just seconds.

Answer 5

Most of the simple programs have computation time in milli-seconds. So, i suppose, you will find this useful.
#include <time.h>
#include <stdio.h>
int main(){
clock_t start = clock();
// Execuatable code
clock_t stop = clock();
double elapsed = (double)(stop - start) * 1000.0 / CLOCKS_PER_SEC;
printf("Time elapsed in ms: %f", elapsed);
}
If you want to compute the runtime of the entire program and you are on a Unix system, run your program using the time command like this time ./a.out

Answer 6

(All answers here are lacking, if your sysadmin changes the systemtime, or your timezone has differing winter- and sommer-times. Therefore...)
On linux use: clock_gettime(CLOCK_MONOTONIC_RAW, &time_variable);
It's not affected if the system-admin changes the time, or you live in a country with winter-time different from summer-time, etc.
#include <stdio.h>
#include <time.h>
#include <unistd.h> /* for sleep() */
int main() {
struct timespec begin, end;
clock_gettime(CLOCK_MONOTONIC_RAW, &begin);
sleep(1); // waste some time
clock_gettime(CLOCK_MONOTONIC_RAW, &end);
printf ("Total time = %f seconds\n",
(end.tv_nsec - begin.tv_nsec) / 1000000000.0 +
(end.tv_sec - begin.tv_sec));
}
man clock_gettime states:
CLOCK_MONOTONIC
Clock that cannot be set and represents monotonic time since some unspecified starting point. This clock is not affected by discontinuous jumps in the system time
(e.g., if the system administrator manually changes the clock), but is affected by the incremental adjustments performed by adjtime(3) and NTP.

Answer 7

Thomas Pornin's answer as macros:
#define TICK(X) clock_t X = clock()
#define TOCK(X) printf("time %s: %g sec.\n", (#X), (double)(clock() - (X)) / CLOCKS_PER_SEC)
Use it like this:
TICK(TIME_A);
functionA();
TOCK(TIME_A);
TICK(TIME_B);
functionB();
TOCK(TIME_B);
Output:
time TIME_A: 0.001652 sec.
time TIME_B: 0.004028 sec.

Answer 8

A lot of answers have been suggesting clock() and then CLOCKS_PER_SEC from time.h. This is probably a bad idea, because this is what my /bits/time.h file says:
/* ISO/IEC 9899:1990 7.12.1: <time.h>
The macro `CLOCKS_PER_SEC' is the number per second of the value
returned by the `clock' function. */
/* CAE XSH, Issue 4, Version 2: <time.h>
The value of CLOCKS_PER_SEC is required to be 1 million on all
XSI-conformant systems. */
# define CLOCKS_PER_SEC 1000000l
# if !defined __STRICT_ANSI__ && !defined __USE_XOPEN2K
/* Even though CLOCKS_PER_SEC has such a strange value CLK_TCK
presents the real value for clock ticks per second for the system. */
# include <bits/types.h>
extern long int __sysconf (int);
# define CLK_TCK ((__clock_t) __sysconf (2)) /* 2 is _SC_CLK_TCK */
# endif
So CLOCKS_PER_SEC might be defined as 1000000, depending on what options you use to compile, and thus it does not seem like a good solution.

Answer 9

#include<time.h>
#include<stdio.h>
int main(){
clock_t begin=clock();
int i;
for(i=0;i<100000;i++){
printf("%d",i);
}
clock_t end=clock();
printf("Time taken:%lf",(double)(end-begin)/CLOCKS_PER_SEC);
}
This program will work like charm.

Answer 10

You have to take into account that measuring the time that took a program to execute depends a lot on the load that the machine has in that specific moment.
Knowing that, the way of obtain the current time in C can be achieved in different ways, an easier one is:
#include <time.h>
#define CPU_TIME (getrusage(RUSAGE_SELF,&ruse), ruse.ru_utime.tv_sec + \
ruse.ru_stime.tv_sec + 1e-6 * \
(ruse.ru_utime.tv_usec + ruse.ru_stime.tv_usec))
int main(void) {
time_t start, end;
double first, second;
// Save user and CPU start time
time(&start);
first = CPU_TIME;
// Perform operations
...
// Save end time
time(&end);
second = CPU_TIME;
printf("cpu : %.2f secs\n", second - first);
printf("user : %d secs\n", (int)(end - start));
}
Hope it helps.
Regards!

Answer 11

ANSI C only specifies second precision time functions. However, if you are running in a POSIX environment you can use the gettimeofday() function that provides microseconds resolution of time passed since the UNIX Epoch.
As a side note, I wouldn't recommend using clock() since it is badly implemented on many(if not all?) systems and not accurate, besides the fact that it only refers to how long your program has spent on the CPU and not the total lifetime of the program, which according to your question is what I assume you would like to measure.

Answer 12

I've found that the usual clock(), everyone recommends here, for some reason deviates wildly from run to run, even for static code without any side effects, like drawing to screen or reading files. It could be because CPU changes power consumption modes, OS giving different priorities, etc...
So the only way to reliably get the same result every time with clock() is to run the measured code in a loop multiple times (for several minutes), taking precautions to prevent the compiler from optimizing it out: modern compilers can precompute the code without side effects running in a loop, and move it out of the loop., like i.e. using random input for each iteration.
After enough samples are collected into an array, one sorts that array, and takes the middle element, called median. Median is better than average, because it throws away extreme deviations, like say antivirus taking up all CPU up or OS doing some update.
Here is a simple utility to measure execution performance of C/C++ code, averaging the values near median: https://github.com/saniv/gauge
I'm myself still looking for a more robust and faster way to measure code. One could probably try running the code in controlled conditions on bare metal without any OS, but that will give unrealistic result, because in reality OS does get involved.
x86 has these hardware performance counters, which including the actual number of instructions executed, but they are tricky to access without OS help, hard to interpret and have their own issues ( http://archive.gamedev.net/archive/reference/articles/article213.html ). Still they could be helpful investigating the nature of the bottle neck (data access or actual computations on that data).

Answer 13

Every solution's are not working in my system.
I can get using
#include <time.h>
double difftime(time_t time1, time_t time0);

Answer 14

Some might find a different kind of input useful: I was given this method of measuring time as part of a university course on GPGPU-programming with NVidia CUDA (course description). It combines methods seen in earlier posts, and I simply post it because the requirements give it credibility:
unsigned long int elapsed;
struct timeval t_start, t_end, t_diff;
gettimeofday(&t_start, NULL);
// perform computations ...
gettimeofday(&t_end, NULL);
timeval_subtract(&t_diff, &t_end, &t_start);
elapsed = (t_diff.tv_sec*1e6 + t_diff.tv_usec);
printf("GPU version runs in: %lu microsecs\n", elapsed);
I suppose you could multiply with e.g. 1.0 / 1000.0 to get the unit of measurement that suits your needs.

Answer 15

If you program uses GPU or if it uses sleep() then clock() diff gives you smaller than actual duration. It is because clock() returns the number of CPU clock ticks. It only can be used to calculate CPU usage time (CPU load), but not the execution duration. We should not use clock() to calculate duration. We still should use gettimeofday() or clock_gettime() for duration in C.

Answer 16

perf tool is more accurate to be used in order to collect and profile the running program. Use perf stat to show all information related to the program being executed.

Answer 17

As simple as possible by using function-like macro
#include <stdio.h>
#include <time.h>
#define printExecTime(t) printf("Elapsed: %f seconds\n", (double)(clock()-(t)) / CLOCKS_PER_SEC)
int factorialRecursion(int n) {
return n == 1 ? 1 : n * factorialRecursion(n-1);
}
int main()
{
clock_t t = clock();
int j=1;
for(int i=1; i <10; i++ , j*=i);
printExecTime(t);
// compare with recursion factorial
t = clock();
j = factorialRecursion(10);
printExecTime(t);
return 0;
}

Answer 18

Comparison of execution time of bubble sort and selection sort
I have a program which compares the execution time of bubble sort and selection sort.
To find out the time of execution of a block of code compute the time before and after the block by
clock_t start=clock();
…
clock_t end=clock();
CLOCKS_PER_SEC is constant in time.h library
Example code:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main()
{
int a[10000],i,j,min,temp;
for(i=0;i<10000;i++)
{
a[i]=rand()%10000;
}
//The bubble Sort
clock_t start,end;
start=clock();
for(i=0;i<10000;i++)
{
for(j=i+1;j<10000;j++)
{
if(a[i]>a[j])
{
int temp=a[i];
a[i]=a[j];
a[j]=temp;
}
}
}
end=clock();
double extime=(double) (end-start)/CLOCKS_PER_SEC;
printf("\n\tExecution time for the bubble sort is %f seconds\n ",extime);
for(i=0;i<10000;i++)
{
a[i]=rand()%10000;
}
clock_t start1,end1;
start1=clock();
// The Selection Sort
for(i=0;i<10000;i++)
{
min=i;
for(j=i+1;j<10000;j++)
{
if(a[min]>a[j])
{
min=j;
}
}
temp=a[min];
a[min]=a[i];
a[i]=temp;
}
end1=clock();
double extime1=(double) (end1-start1)/CLOCKS_PER_SEC;
printf("\n");
printf("\tExecution time for the selection sort is %f seconds\n\n", extime1);
if(extime1<extime)
printf("\tSelection sort is faster than Bubble sort by %f seconds\n\n", extime - extime1);
else if(extime1>extime)
printf("\tBubble sort is faster than Selection sort by %f seconds\n\n", extime1 - extime);
else
printf("\tBoth algorithms have the same execution time\n\n");
}