Re: [xmoto-dev] svn 1404 : merged benetnash with trunk

[nadenislamarre@xxxxxxx]

benenash :

To avoid this pb,
i wrote this in std::string App::formatTime(float fSecs)
    /* hum, case, in which 0.9800 * 100.0 => 0.9799999*/
    if(((int)(nHres * 100.0)) < ((int)((nHres * 100.0) + 0.001))) {
      nH = ((int)((nHres * 100.0) + 0.001));
      nH %= 100;
    }

however, we should write funtions called :

float timetoFloatTime(int i_minutes, int i_seconds, int i_seconds);
void  timeToIntTime(float i_time, int& o_minutes, int& o_seconds, int&
o_hundreaths);

which assumes :
i_minutes : i_seconds : i_seconds >= float
to FIX definitively this pb into 2 functions.

What do you think about this ?

Nicolas


Quoting Janek Polak <benetnash@xxxxxxxxxxxxxx>:

> Dnia Tue, 1 May 2007, Nicolas Adenis-Lamarre naskrobal
> >  Janek Polak a écrit :
> > > Today morning I've invented that this floating point differences can be
> > > removed by changeing (replay_time < highscore_time) to (replay_time -
> > > highscore_time < -0.01). I'm not perfectly sure if it works, but this
> > > solution looks fine. I left comment in case someone invent something
> > > better.
> >  i found a nice feature of sqlite,
> >  it has not a lot of function, however, it allows to add news !
>
> >   sqlite3_result_double(i_context, (double)((int)(v_value)));
>
> It is not working - it truncates 23.0 to 22.0 and so on - from numbers
> witout decimal digits it substracts 1. I think that issue may be caused
> by some floating point errors. Look at this:
>
> testowe$ cat x.cpp
> #include <iostream>
> using namespace std;
> main()
> {
>
> 	double x = 23.456;
> 	printf("%2.20f\n", x);
> 	printf("%2.20f\n", 100 * x);
> }
> testowe$ ./x
> 23.45599999999999951683
> 2345.59999999999990905053
>
>
> This 23.456 will be truncated by your function to 2345.5
> My nasty workaround is:
>
> sqlite3_result_double(i_context, (double)((int)(v_value + 0.0000001)));
>
> On the other hand we could round number with:
>
> sqlite3_result_double(i_context, floor(v_value + 0.5)));
>
> Anyway I've just commited first workaround and added xm_floor to both
> times:
>
> xm_floor(h.finishTime*100.0) > xm_floor(r.finishTime*100.0))
>
>
>