Rockbox mail archive
Subject: Re: math
Ray Lambert wrote:
> BlueChip wrote:
>
>> while (x>0) r = (r<<3)+(r<<1); // r*8 + r*2
>> ^whoops
>>
>> what values of 'x' ? all positive? include 0?
>>
>> if (x<0) return(0); // cannot handle negative values of x
>> if (x>9) return(1); // result will be > (2^32)1
>> unsigned long r = 1;
>> while (x>0) r = (r<<4)+(r<<1); // r*8 + r*2
>> return (r);
>
> It might be a good idea to decrement x at some point too.... :)
>
>> while (x>0) r = (r<<3)+(r<<1); // r*8 + r*2
>
> perhaps:
>
> for (; x>0; x) r = (r<<3)+(r<<1); // r*8 + r*2
>
> ~ray
And catch x=0. And set your base at 10 not 1 (remember, this is exp10).
if (x == 0) return 1; // 10^0 is always 1
if (x < 0) return 0; // cannot handle negative values of x
if (x > 9) return 1; // result will be > (2^32)1
unsigned long r = 10;
while (x) r = (r << 3) + (r << 1); // r*8 + r*2
return r;
Dave
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Received on Wed Jul 6 22:07:31 2005
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