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Userrr
Joined: 09 Jul 2004 Posts: 40 Location: Europe

float problem 
Posted: Tue Dec 28, 2004 2:47 pm 


Sometimes after devide this:
x unsigned int32;
y float;
z unsigned int32;
x/y=z;
100000/10=9999;  ? Why not 10000; were 1
But sometimes = 10000  That correct !
How correct this if I don't need use printf or sprintf;
and tell me please what mean letter ' L ' after some digits? 


Guest

Re: float problem 
Posted: Wed Dec 29, 2004 12:09 am 


Userrr wrote:  Sometimes after devide this:
x unsigned int32;
y float;
z unsigned int32;
x/y=z;
100000/10=9999;  ? Why not 10000; were 1
But sometimes = 10000  That correct !
How correct this if I don't need use printf or sprintf;
and tell me please what mean letter ' L ' after some digits? 
try this
z = x / y *(1+1e7);
then you should have 10000 instead of 9999.999 


Ttelmah Guest


Posted: Wed Dec 29, 2004 6:07 am 


'L', is a signal to the compiler, that this value is to be treated as a 'Long'. Though generally, the compiler will convert types, the 'casting' (automatic conversion), in the CCS C, is sometimes a little'lacking', and it can therefore be worthwhile 'forcing' the type.
On the first question about division, what you show, should work. However there is a major 'caveat', if you are dealing with real numbers, rather than the fixed values shown. If (for instance), the result of some aritmetic, is 99999.9921875 (the next value 'below' 100000, that is actually representable by the floating format), and you then calculate this /10, and convert this to an integer (which is what is happening in your z=x/y calculation  the sum as you show it is 'reversed'  you cannot put a value 'into' a calculation), this will be rounded down in the conversion, and will result in 9999. Also 'beware' that CCS, does not reliably cast forward (converting parts of a sum to the higher arithmetic type), but can end up doing the arithmetic in the 'lower' type (int32).
Hence if you want to convert types like this, it is better to explicitly cast, and also to force 4/5 rounding if this is what is wanted.
So a line like:
z=x/(int32)(y+0.49999);
Will force 'y' to be converted to an integer before the aritmetic, and be rounded 4/5, when this happens, and may give what you expect.
Best Wishes 




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