Why CPU time is negative

Question

I am trying to measure the CPU time of following code - struct timespec time1, time2, temp_time;

          clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &time1);

          long diff = 0;

          for(int y=0; y<n; y++) {

                for(int x=0; x<n; x++) {

                float v = 0.0f;

                 for(int i=0; i<n; i++)

                     v += a[y * n + i] * b[i * n + x];

                       c[y * n + x] = v;

                   }

            }
        clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &time2);

        temp_time.tv_sec = time2.tv_sec - time1.tv_sec;

        temp_time.tv_nsec = time2.tv_nsec - time1.tv_nsec;

        diff = temp_time.tv_sec * 1000000000 + temp_time.tv_nsec; 

       printf("finished calculations using CPU in %ld ms \n", (double) diff/1000000); 

But the time value is negative when i increase the value of n. Code prints correct value for n = 500 but it prints negative value for n = 700 Any help would be appreciated.

Here is the full code structure -

void run(float A[], float B[], float C[], int nelements){
    struct timespec time1, time2, temp_time;

          clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &time1);

          long diff = 0;

          for(int y=0; y<nelements; y++) {

                for(int x=0; x<nelements; x++) {

                float v = 0.0f;

                 for(int i=0; i<nelements; i++)

                     v += A[y * nelements + i] * B[i * nelements + x];

                       C[y * nelements + x] = v;

                   }

            }
        clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &time2);

        temp_time.tv_sec = time2.tv_sec - time1.tv_sec;

        temp_time.tv_nsec = time2.tv_nsec - time1.tv_nsec;

        diff = temp_time.tv_sec * 1000000000 + temp_time.tv_nsec; 

       printf("finished calculations using CPU in %ld ms \n"(double) diff/1000000); 
}

This function abovr is called from different fil as follows:

SIZE = 500;

a = (float*)malloc(SIZE * SIZE * sizeof(float));

b = (float*)malloc(SIZE * SIZE * sizeof(float));

c = (float*)malloc(SIZE * SIZE * sizeof(float));

//initialize a &b
run(&a[SIZE],&b[SIZE],&c[SIZE],SIZE);

Answer 1

看起来像溢出使用unsigned long或更好的double用于diff

Answer 2

The 'tv_nsec' field should never exceed 10^9 (1000000000), for obvious reasons:

if (time1.tv_nsec < time2.tv_nsec)
{
    int adj = (time2.tv_nsec - time1.tv_nsec) / (1000000000) + 1;

    time2.tv_nsec -= (1000000000) * adj;
    time2.tv_sec += adj;
}

if (time1.tv_nsec - time2.tv_nsec > (1000000000))
{
    int adj = (time1.tv_nsec - time2.tv_nsec) / (1000000000);

    time2.tv_nsec += (1000000000) * adj;
    time2.tv_sec -= adj;
}

temp_time.tv_sec = time1.tv_sec - time2.tv_sec;
temp_time.tv_nsec = time1.tv_nsec - time2.tv_nsec;

diff = temp_time.tv_sec * (1000000000) + temp_time.tv_nsec;

This code could be simplified, as it makes no assumptions about the sign of the 'tv_sec' field. Most Linux sys headers (and glibc?) provide macros to handle this sort of timespec arithmetic correctly don't they?

Answer 3

One of possible problem causes is that the printf format is for a long signed integer value ( %ld ), but the parameter has the double type. To fix the problem is necessary change %ld to %lf in the format string.

Answer 4

Look at your print statement:

printf("finished calculations using CPU in %ld ms \n", (double) diff/1000000);

The second parameter you pass is a double, but you are printing out this floating point value as a long (%ld). I suspect that's half your problem.

This may generate better results:

printf("finished calculations using CPU in %f ms \n", diff/1000000.0);

I also agree with keety, you likely should be using unsigned types Or you could possibly avoid the overflow issues altogether by staying in millisecond units instead of nanoseconds. Here's why I stick with 64-bit unsigned integers and just stay in the millisecond realm.

unsigned long long diffMilliseconds;

diffMilliseconds = (time2.tv_sec * 1000LL + time2.tv_nsec/1000000) - (time1.tv_sec * 1000LL + time1.tv_nsec/1000000);

printf("finished calculations using CPU in %llu ms \n", diffMilliseconds);