We value your feedback.
Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!
Course Of The Month
7 days, 22 hours left to enrollBecome a Premium Member and unlock a new, free course in leading technologies each month.
Solved
Problem timing a function in C
Posted on 2007-11-27
So in my code below I loop doing:
1) create some data (the exact same data is created every time), and put it into a double array.
2) record clock time
3) process the data
4) calculate the elapsed time, and add it to the elapsed total
5) print out the running total
The first 50 or so times through the loop the elapsed time prints out as 0.0000000.....
Then all of a sudden it jumps to:
19991590273549713308756725
and stays there for the remainder of the 100 iterations, even though all 100 of the iterations take the same amount of time (or at least with a very small degree of variation).
Can anybody tell me what is up with this? I'm trying to get an average for how long the function takes to run.
Thanks
1) create some data (the exact same data is created every time), and put it into a double array.
2) record clock time
3) process the data
4) calculate the elapsed time, and add it to the elapsed total
5) print out the running total
The first 50 or so times through the loop the elapsed time prints out as 0.0000000.....
Then all of a sudden it jumps to:
19991590273549713308756725
and stays there for the remainder of the 100 iterations, even though all 100 of the iterations take the same amount of time (or at least with a very small degree of variation).
Can anybody tell me what is up with this? I'm trying to get an average for how long the function takes to run.
Thanks
int main()
{
int i = 0;
int numIterations = 100;
double elapsed;
clock_t before;
//complexData is a double []
while(i < numIterations)
{
createSomeData(complexData, SIZE);
before = clock();
processData(complexData-1, SIZE, 1);
elapsed = elapsed + ((clock()-before)/CLOCKS_PER_SEC);
printf("Time: %f\n", 1, (elapsed));
i++;
}
printf("Avg Time: %.6f\n", 1, (elapsed)/(double)numIterations);
return 0;
}
[X]
Welcome to Experts Exchange
Add your voice to the tech community where 5M+ people just like you are talking about what matters.
- Help others & share knowledge
- Earn cash & points
- Learn & ask questions
7
2-
2
- +3
15 Comments
Your code looks good and an approximation of it works for me with VS2005 and Windows.
I suggest trying an experiment:
1) Add a sleep function, such as
#include <time.h>
void sleep(unsigned int mseconds)
{
clock_t goal = mseconds + clock();
while (goal > clock());
}
2) Comment out the body of your two functions. Insert a call to sleep(100), replacing the commented-out function bodies.
3) Run the program again. Do you still have the problem?
I suggest trying an experiment:
1) Add a sleep function, such as
#include <time.h>
void sleep(unsigned int mseconds)
{
clock_t goal = mseconds + clock();
while (goal > clock());
}
2) Comment out the body of your two functions. Insert a call to sleep(100), replacing the commented-out function bodies.
3) Run the program again. Do you still have the problem?
jaime,
>>> it should be: clock_t before = clock();
True, but he initializes it before he uses it, so that can't be the problem
>>> elapsed is not initialized, should be: double elapsed = 0.0;
True, but his first 50 results are 0.0, so I assume he's running in debug and it has been initialized for him
I don't see that either of these issues (which *do* need to be fixed) are the issue at hand.
Joe
>>> it should be: clock_t before = clock();
True, but he initializes it before he uses it, so that can't be the problem
>>> elapsed is not initialized, should be: double elapsed = 0.0;
True, but his first 50 results are 0.0, so I assume he's running in debug and it has been initialized for him
I don't see that either of these issues (which *do* need to be fixed) are the issue at hand.
Joe
When I do as you suggested (with the sleep) here is what I get:
Total Time So Far: 19991590273549713308756725
Total Time So Far: 19991590273549713308756725
Total Time So Far: -8679222230972213701681168
Total Time So Far: 19991590273549713308756725
Total Time So Far: -0.000000
With the code below.
Whats going on!?
Total Time So Far: 19991590273549713308756725
Total Time So Far: 19991590273549713308756725
Total Time So Far: -8679222230972213701681168
Total Time So Far: 19991590273549713308756725
Total Time So Far: -0.000000
With the code below.
Whats going on!?
int main()
{
int i = 0;
int numIterations = 5;
double elapsed = 0.0;
clock_t before;
clock_t goal;
while(i < numIterations)
{
before = clock();
clock_t goal = 100 + clock();
while(goal > clock())
{
}
elapsed = elapsed + (((double)clock()-
(double)before)/CLOCKS_PER_SEC);
printf("Total Time So Far: %f\n", 1, (elapsed));
i++;
}
return 0;
}
printf("Total Time So Far: %f\n", 1, (elapsed));
should be
printf("Total Time So Far: %i, %f\n", 1, (elapsed));
should be
printf("Total Time So Far: %i, %f\n", 1, (elapsed));
>> can you try
>> elapsed = elapsed + (((double)clock()-before)/
Did you check ozo's post ?
That's pretty sure to be the problem, since your original code was doing integer division, and you want it to do double division. I would do this however :
elapsed += ((double) (clock() - before)) / CLOCKS_PER_SEC;
You also need to initialize elapsed to 0 of course as jaime_olivares suggested :
elapsed = 0.0
And the printf statement needs to be fixed as jaime_olivares also said :
printf("Time: %f\n", elapsed);
>> elapsed = elapsed + (((double)clock()-before)/
Did you check ozo's post ?
That's pretty sure to be the problem, since your original code was doing integer division, and you want it to do double division. I would do this however :
elapsed += ((double) (clock() - before)) / CLOCKS_PER_SEC;
You also need to initialize elapsed to 0 of course as jaime_olivares suggested :
elapsed = 0.0
And the printf statement needs to be fixed as jaime_olivares also said :
printf("Time: %f\n", elapsed);
>> the printf problem is all !!!
The other two I mentioned (and ozo and you before me) are problems too ;)
The other two I mentioned (and ozo and you before me) are problems too ;)
What's happening is you have an extra "1," in the printf list. This causes printf to try to print out as a double the value of an integer "1" in the lower 32 bits followed by the lower 32 bits of the elapsed time as the upper 32 bits of a double. So you're going to see huge random-looking numbers.
You see printf has no way at run-time or compile time to check for type-compatibility of the %format specifiers versus the actual parameters. This is exacerbated by C's silent promotion of all parameters to ints and doubles.
Another problem is that your clock() function on your computer probably has a granularity of many milliseconds. Depending on the time spent in the timed function you might have to bump up the loop repetition count so that clock()'s granularity is not so noticeable.
Also keep in mind that repeating a block of code with the same input data is a misleading way to time code. Most computers have code and data caches so when you repeat code and data it may run *many* times faster than it will in a real life situation.
You see printf has no way at run-time or compile time to check for type-compatibility of the %format specifiers versus the actual parameters. This is exacerbated by C's silent promotion of all parameters to ints and doubles.
Another problem is that your clock() function on your computer probably has a granularity of many milliseconds. Depending on the time spent in the timed function you might have to bump up the loop repetition count so that clock()'s granularity is not so noticeable.
Also keep in mind that repeating a block of code with the same input data is a misleading way to time code. Most computers have code and data caches so when you repeat code and data it may run *many* times faster than it will in a real life situation.
Featured Post
We value your feedback.
Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!
Question has a verified solution.
If you are experiencing a similar issue, please ask a related question
An Outlet in Cocoa is a persistent reference to a GUI control; it connects a property (a variable) to a control. For example, it is common to create an Outlet for the text field GUI control and change the text that appears in this field via that Ou…
Summary:
This tutorial covers some basics of pointer, pointer arithmetic and function pointer.
What is a pointer:
A pointer is a variable which holds an address. This address might be address of another variable/address of devices/address of fu…
The goal of this video is to provide viewers with basic examples to understand and use structures in the C programming language.
The goal of this video is to provide viewers with basic examples to understand recursion in the C programming language.
Suggested Courses
Course of the Month7 days, 22 hours left to enroll
728 members asked questions and received personalized solutions in the past 7 days.
Join the community of 500,000 technology professionals and ask your questions.