Re[2]: [Dev-C++] POW question

[andrey <ak...@bk...>]

(2^8+1)/ 2  isn't pow(2,8)/2+1

try 
double test = (pow(2.0,8.0)+1.0)/2.0;

it will give you 128.5 because compiler know each value in expression DOUBLE
(sorry for my english)

Andrey

> 
> >I'm not sure how abs() would help.. unless that's to ensure that the
> >original value is positive before multiplying by negative one?
> >-Lukas
> 
> This is still not working right,
> Iv'e been scratching my head on this all day..
> 
> I'm trying to get the value of 128.5
> But instead all the mentioned formula's are giving me 129...
> 
> What i'm trying to express is this..
> 
> (2^8+1)/ 2 = 128.5
> 
> So in C/C++ it should be pow(2,8)/2+1  (However it returns 129)
> 
> Here is a quick test....
> ----------------------------------------------------------
> #include <stdio.h>
> #include <math.h>
> void main()
> {
> double test;
> test=pow(2,8)/2+1; //figure 2 to the eighth power +1
> printf("The number is %0.0f \n",test);
> system("PAUSE");
> return 0;
> }
> -----------------------------------------------------------
> #include <stdio.h>
> #include <math.h>
> void main()
> {
> float test;
> test=((2^8+1) / 2); //figure 2 to the eighth power +1
> printf("The number is %0.0f \n",test);
> system("PAUSE");
> return 0;
> }
> -------------------------------------------------------------
> 
> Here is a quick snippett from the actual source code i'm using to position 
> points.
> 
> sprintf( cmd, "MAKEBOX <%g %g %g> <%g %g %f> <%d %d %d>",
>       -128.5, -0.0, -128.5,  pow(2,8)/2+1, 0.0, pow(2,8)/2+1, 1, 1, 1 );
>    local->evaluate( local->data, cmd );
> 
> Both suffer from the returning the same 129 figure....
> I must be missing something stupid...  (i'll try again in the morning)
> 
> Thanks for the help guys..
> 
> Cheers,
> Scott. 
> 
> 
> 
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