Which is faster in these fread() commands?

archieval · 05-05-2010, 08:30 PM

Which is faster:

fread(&buf, 1, 1024, fp);

or

fread(&buf, 1024, 1, fp);

?

Regards,
archieval

Sergei Steshenko · 05-05-2010, 08:54 PM

Quote:

Originally Posted by archieval

Which is faster:

fread(&buf, 1, 1024, fp);

or

fread(&buf, 1024, 1, fp);

?

Regards,
archieval

Probably the one with greater item size, but why won't you check it yourself ?

vikas027 · 05-05-2010, 08:58 PM

You can use time command as below.

Code:

time fread(&buf, 1, 1024, fp);

Sergei Steshenko · 05-05-2010, 08:59 PM

Quote:

Originally Posted by vikas027

You can use time command as below.

Code:

time fread(&buf, 1, 1024, fp);

Do you think your suggestion is a valid "C" code ?

archieval · 05-05-2010, 09:34 PM

time is a unix system command. I used the codes below to test but I got seg fault...

Code:

#include <stdlib.h>
#include <stdio.h>
#include <time.h>
#include <sys/time.h>

int main(void)
{
        FILE *fp;
        char *buf;
        unsigned long t1, t2;
        struct tm tm1, tm2;
        struct timeval tb1, tb2, tb3, tb4;

        fp = fopen("../hi_archie/dvr_test.c", "r");
        if (fp == NULL)
        {
                printf("error\n");
                return 1;
        }
        
        buf = malloc(2048);
        if(buf == NULL)
                return;
        printf("xxx\n");

        gettimeofday(&tb1, NULL);
        fread(&buf, 1024, 1, fp);
        gettimeofday(&tb2, NULL);

        printf("time lapsed %lu\n", tb2.tv_usec - tb1.tv_usec);

        fseek(fp, 0, SEEK_SET);
        gettimeofday(&tb3, NULL);
        fread(&buf, 1, 1024, fp);
        gettimeofday(&tb4, NULL);
        printf("time lapsed %lu\n", tb4.tv_usec - tb3.tv_usec);

        free(buf);

        return 0;
}

result:
$ gcc -Wall test.c ; ./a.out

test.c: In function ‘main’:
test.c:23: warning: ‘return’ with no value, in function returning non-void
test.c:11: warning: unused variable ‘tm2’
test.c:11: warning: unused variable ‘tm1’
test.c:10: warning: unused variable ‘t2’
test.c:10: warning: unused variable ‘t1’
time lapsed 18
Segmentation fault

Sergei Steshenko · 05-05-2010, 09:57 PM

Quote:

Originally Posted by archieval

time is a unix system command. I used the codes below to test but I got seg fault...

Code:

#include <stdlib.h>
#include <stdio.h>
#include <time.h>
#include <sys/time.h>

int main(void)
{
        FILE *fp;
        char *buf;
        unsigned long t1, t2;
        struct tm tm1, tm2;
        struct timeval tb1, tb2, tb3, tb4;

        fp = fopen("../hi_archie/dvr_test.c", "r");
        if (fp == NULL)
        {
                printf("error\n");
                return 1;
        }
        
        buf = malloc(2048);
        if(buf == NULL)
                return;
        printf("xxx\n");

        gettimeofday(&tb1, NULL);
        fread(&buf, 1024, 1, fp);
        gettimeofday(&tb2, NULL);

        printf("time lapsed %lu\n", tb2.tv_usec - tb1.tv_usec);

        fseek(fp, 0, SEEK_SET);
        gettimeofday(&tb3, NULL);
        fread(&buf, 1, 1024, fp);
        gettimeofday(&tb4, NULL);
        printf("time lapsed %lu\n", tb4.tv_usec - tb3.tv_usec);

        free(buf);

        return 0;
}

result:
$ gcc -Wall test.c ; ./a.out

test.c: In function ‘main’:
test.c:23: warning: ‘return’ with no value, in function returning non-void
test.c:11: warning: unused variable ‘tm2’
test.c:11: warning: unused variable ‘tm1’
test.c:10: warning: unused variable ‘t2’
test.c:10: warning: unused variable ‘t1’
time lapsed 18
Segmentation fault

So, first get rid of warnings, the determine the line causing segfault.

...

Use stderr, not stdout, for diagnostic output.

archieval · 05-05-2010, 11:18 PM

wrong in dereferencing buf >___<

should be

fread(buf, 1024, 1, fp);
fread(buf, 1, 1024, fp);

its ok now.

time lapsed 30
time lapsed 6

so the second is the faster one.

Sergei Steshenko · 05-06-2010, 12:20 AM

Quote:

Originally Posted by archieval

wrong in dereferencing buf >___<

should be

fread(buf, 1024, 1, fp);
fread(buf, 1, 1024, fp);

its ok now.

time lapsed 30
time lapsed 6

so the second is the faster one.

I wouldn't be so sure - run your test a number of times.

broken · 05-06-2010, 01:09 AM

How about using the "correct" one rather than the fastest one? Good luck handling the return value if you pick the incorrect version because it is faster.

archieval · 05-06-2010, 03:04 AM

[archieval@localhost hi_archie_empty]$ ./a.out
time lapsed 30
time lapsed 5
[archieval@localhost hi_archie_empty]$ ./a.out
time lapsed 32
time lapsed 5
[archieval@localhost hi_archie_empty]$ ./a.out
time lapsed 20
time lapsed 3
[archieval@localhost hi_archie_empty]$ ./a.out
time lapsed 30
time lapsed 5
[archieval@localhost hi_archie_empty]$ ./a.out
time lapsed 30
time lapsed 6
[archieval@localhost hi_archie_empty]$ ./a.out
time lapsed 28
time lapsed 5
[archieval@localhost hi_archie_empty]$ ./a.out
time lapsed 30
time lapsed 5
[archieval@localhost hi_archie_empty]$ ./a.out
time lapsed 30
time lapsed 6
[archieval@localhost hi_archie_empty]$ ./a.out
time lapsed 33
time lapsed 5
[archieval@localhost hi_archie_empty]$ ./a.out
time lapsed 31
time lapsed 6
[archieval@localhost hi_archie_empty]$ ./a.out
time lapsed 50
time lapsed 6
[archieval@localhost hi_archie_empty]$ ./a.out
time lapsed 35
time lapsed 6
[archieval@localhost hi_archie_empty]$ ./a.out
time lapsed 31
time lapsed 7
[archieval@localhost hi_archie_empty]$ ./a.out
time lapsed 43
time lapsed 5
[archieval@localhost hi_archie_empty]$ ./a.out
time lapsed 30
time lapsed 5
[archieval@localhost hi_archie_empty]$ ./a.out
time lapsed 32
time lapsed 6
[archieval@localhost hi_archie_empty]$ ./a.out
time lapsed 30
time lapsed 6

archieval · 05-06-2010, 03:14 AM

i changed the code:

Code:

        for (i = 0; i<100; i++)
        {
                gettimeofday(&tb1, NULL);
                fread(buf, 1024, 1, fp);
                gettimeofday(&tb2, NULL);

                printf("%lu, ", tb2.tv_usec - tb1.tv_usec);

                gettimeofday(&tb3, NULL);
                fread(buf, 1, 1024, fp);
                gettimeofday(&tb4, NULL);
                printf("%lu\n", tb4.tv_usec - tb3.tv_usec);

        }

result:

33, 1
1, 1
8, 1
2, 1
7, 2
1, 2
5, 1
1, 1
5, 1
1, 1
5, 1
1, 2
5, 1
1, 1
5, 1
2, 1
5, 1
1, 1
5, 2
2, 1
5, 1
1, 2
6, 2
1, 2
5, 2
2, 2
5, 1
1, 2
6, 1
1, 2
6, 1
1, 1
6, 1
2, 1
6, 1
1, 1
5, 1
2, 1
6, 2
1, 1
5, 2
1, 1
5, 1
1, 2
6, 1
1, 1
5, 1
1, 2
6, 1
2, 1
7, 1
2, 1
6, 1
1, 1
5, 2
1, 1
7, 2
1, 1
5, 2
2, 1
5, 1
1, 1
7, 2
1, 2
6, 1
1, 2
5, 2
2, 1
5, 1
2, 1
5, 2
2, 1
6, 2
2, 2
6, 1
1, 1
5, 1
1, 1
5, 2
1, 1
6, 2
1, 2
6, 1
2, 1
5, 2
1, 2
6, 2
2, 1
6, 2
2, 1
5, 1
2, 1
5, 1
1, 2
5, 1
1, 1
5, 2
2, 2
5, 1
1, 1

Sergei Steshenko · 05-06-2010, 10:57 AM

Quote:

Originally Posted by archieval

i changed the code:

Code:

        for (i = 0; i<100; i++)
        {
                gettimeofday(&tb1, NULL);
                fread(buf, 1024, 1, fp);
                gettimeofday(&tb2, NULL);

                printf("%lu, ", tb2.tv_usec - tb1.tv_usec);

                gettimeofday(&tb3, NULL);
                fread(buf, 1, 1024, fp);
                gettimeofday(&tb4, NULL);
                printf("%lu\n", tb4.tv_usec - tb3.tv_usec);

        }

result:

33, 1
1, 1
8, 1
2, 1
7, 2
1, 2
5, 1
1, 1
5, 1
1, 1
5, 1
1, 2
5, 1
1, 1
5, 1
2, 1
5, 1
1, 1
5, 2
2, 1
5, 1
1, 2
6, 2
1, 2
5, 2
2, 2
5, 1
1, 2
6, 1
1, 2
6, 1
1, 1
6, 1
2, 1
6, 1
1, 1
5, 1
2, 1
6, 2
1, 1
5, 2
1, 1
5, 1
1, 2
6, 1
1, 1
5, 1
1, 2
6, 1
2, 1
7, 1
2, 1
6, 1
1, 1
5, 2
1, 1
7, 2
1, 1
5, 2
2, 1
5, 1
1, 1
7, 2
1, 2
6, 1
1, 2
5, 2
2, 1
5, 1
2, 1
5, 2
2, 1
6, 2
2, 2
6, 1
1, 1
5, 1
1, 1
5, 2
1, 1
6, 2
1, 2
6, 1
2, 1
5, 2
1, 2
6, 2
2, 1
6, 2
2, 1
5, 1
2, 1
5, 1
1, 2
5, 1
1, 1
5, 2
2, 2
5, 1
1, 1

Now create two test programs - in one you'll have this order:

fread(buf, 1024, 1, fp);
fread(buf, 1, 1024, fp);

and in the other the opposite order:

fread(buf, 1, 1024, fp);
fread(buf, 1024, 1, fp);
.

And compare results. By the way, you already have quite a lot of food for thought.

archieval · 05-06-2010, 08:37 PM

Code:

                gettimeofday(&tb1, NULL);
                fread(buf, 1, 1024, fp);
                gettimeofday(&tb2, NULL);

                printf("%lu, ", tb2.tv_usec - tb1.tv_usec);

                gettimeofday(&tb3, NULL);
                fread(buf, 1024, 1, fp);
                gettimeofday(&tb4, NULL);
                printf("%lu\n", tb4.tv_usec - tb3.tv_usec);

here is the result:

there is some kind of overhead on the first call or at the start of the loop...

Code:

Sergei Steshenko · 05-06-2010, 08:48 PM

Quote:

Originally Posted by archieval

Code:

                gettimeofday(&tb1, NULL);
                fread(buf, 1, 1024, fp);
                gettimeofday(&tb2, NULL);

                printf("%lu, ", tb2.tv_usec - tb1.tv_usec);

                gettimeofday(&tb3, NULL);
                fread(buf, 1024, 1, fp);
                gettimeofday(&tb4, NULL);
                printf("%lu\n", tb4.tv_usec - tb3.tv_usec);

here is the result:

there is some kind of overhead on the first call or at the start of the loop...

Code:

So, seeing all this, do you realize your original question actually doesn't make sense ?

archieval · 05-06-2010, 11:59 PM

are you implying that they are of equal speed?