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kas1e
timemeasure function give 0ms ?
Posted on: 2018/12/5 10:17    #1
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There is test code:

// ppc-amigaos-g++ -D__USE_OLDTIMEVAL__ test_timemeasuer.c

#include <stdio.h>
#include <proto/dos.h>

unsigned int GetTimeStamp()
{
    timeval tNow;
    gettimeofday(&tNow, NULL);
    return ((double)tNow.tv_sec)*1000.0+((double)tNow.tv_usec)/1000.0;
}



int main()
{

int nStartTime=GetTimeStamp();

IDOS->Delay(500);

printf("test time %d ms",GetTimeStamp()-nStartTime);

}


I compile it like

Quote:

ppc-amigaos-g++ -D__USE_OLDTIMEVAL__ test_timemeasuer.c


On running, it wait for some time when Delay() call happens, and then, give me in return 0 ms in printf.

Why ?
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thomas
Re: timemeasure function give 0ms ?
Posted on: 2018/12/5 12:47    #2
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Probably because your time stamp calculation overflows. Your GetTimeStamp() function always returns -1.

It works better if you use (tNow.tv_sec % 86400), but might give wrong results around midnight.
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Ami603
Re: timemeasure function give 0ms ?
Posted on: 2018/12/5 13:01    #3
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@thomas

I would say he's returning seconds and the wait is less than one

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Thematic
Re: timemeasure function give 0ms ?
Posted on: 2018/12/5 13:23    #4
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@Ami603

See AutoDocs/dos - it is ten seconds.
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kas1e
Re: timemeasure function give 0ms ?
Posted on: 2018/12/5 14:32    #5
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@Ami604,Thematic
Yeah it 10 seconds

@Thomas

Once i get rid of double computation, like replacing return string on that:

Quote:

return tNow.tv_sec*1000+tNow.tv_usec/1000;


Then all works. But that code (with double), for sure works on linux and win32.
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jaokim
Re: timemeasure function give 0ms ?
Posted on: 2018/12/5 17:21    #6
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@kas1e

Does ((double)tNow.tv_sec)) really convert from whatever type tv_sec is, to a proper double? Is it perhaps just casting whatever is there and trying to interpret is as a double?

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