Float2Sci

A script for passing floats through strings without loosing precision. wrote this ages ago, it isn't very fast but it is accurate.

Works flawlessly and LSL can parse directly to floats again without any special code :-) just use a (float) typecast.

Changes:

• Cannot assume that optical will work for numbers with large magnitudes. This was necessitated by SCR-397

<lsl> string Float2Sci(float input) {

   if(input == 0.0)//handles negitive zero
       return llDeleteSubString((string)input, -5, -1);//trim off the trailing zero's, don't need them.
   
   float frac = llFabs(input);//we put the negitive back at the end.
   string mantissa = (string)frac;//this may be a string of about 47 characters long.
   integer exponent = -6;//default exponent for optical method
   if(frac == (float)mantissa)
   {
       //optical will work, so all we need do is remove the decimal point and jump to optical.
       mantissa = llDeleteSubString(mantissa, -7, -7);
       }
   else
   {
       //Ugly Math version; ugly in the sense that it is slow and not as elegant as working with it as a string.
       //A) calculate the exponent via approximation of C log2()
       //B) use kludge to avert fatal error in approximation of log2 result (only a problem with values >= 0x1.FFFFF8p127)
       //       the exponent is sometimes reported as 128, which will bork float math, so we subtract the test for 128.
       //       max_float btw is 0x1.FFFFFEp127, so we are only talking a very small number of numbers.
       //C) normalize the float with questionable exponent
       //D) calculate rounding error left from log2 approximation and add to normalization value.
       //       the '|' acts like a '+' in this instance but saves us one byte.
       integer position = (24 | (3 <= frac)) - (integer)( //D
               frac /= (float)("0x1p"+(string)( //C
               exponent = (exponent - (( //B
                   exponent = llFloor(llLog(frac) / 0.69314718055994530941723212145818) //A
               ) == 128))
               ))
               );
   
       //this pushes the float into the interger buffer exactly.
       //since the shift is within integer range, we don't need to make a float.
       integer int = (integer)(frac * (1 << position));
       integer target = (integer)(frac = 0.0);//since the float is in the integer buffer, we need to clear the float buffer.
       
       //we don't use a traditional while loop, and instead opt for a do-while, because it's faster
       //since we may have to do about 128 iteration, this savings is important,
       //the exponent needs one final adjustment because of the shift, we do it here to save memory & it's fasfter.
       
       //The two loops try to make exponent == position by shifting and multiplying.
       //when they are equal, then this should be true ((int * llPow(10, exponent)) == llFabs(input))
       //That is of course assuming that the llPow(10, exponenet) result has enough percision.
       
       //We recycle position for these loops as a temporary buffer. This is so we can save a few operations.
       //If we didn't, then we could actualy optimize the variable out of the code; though it would be slower.
       if(target > (exponent -= position))
       {//apply the rest of the bit shift if |input| < 1
           do
           {
               if(int < 0x19999999)//(0x80000000 / 5)
               {//won't overflow, multiply in 5
                   int = int * 5 + (position = (integer)(frac *= 5.0));
                   frac -= (float)position;
                   target = ~-target;
               }
               else
               {//overflow predicted, devide by 2
                   frac = (frac + (int & 1))/2;
                   int = int >> 1;
                   exponent = -~exponent;
               }
           }while(target ^ exponent);
       }
       else if(target ^ exponent)//target < exponent
       {//apply the rest of the bit shift if |input| > 1
           do
           {
               if(int < 0x40000000) //(0x80000000 / 2)
               {//won't overflow, multiply in 2
                   int = (int << 1) + (position = (integer)(frac *= 2.0));
                   frac -= (float)position;
                   exponent = ~-exponent;
               }
               else
               {//overflow predicted, divide by 5
                   frac = (frac + int%5) / 5.0;
                   int /= 5;
                   target = -~target;
               }
           }while(target ^ exponent);
       }
       //int is now properly calculated, it holds enough data to accurately describe the input in conjunction with exponent.
       //we feed this through optical to clean up the answer.
       mantissa = (string)int;
   }
   //it's not an issue that we may be jumping over the initialization of some of the variables,
   //we initialize everything we use here.
   
   //to accurately describe a float you only need 9 decimal places; so we throw the extra's away
   if(9 < (target = position = llStringLength(mantissa)))
       position = 9;
   //chop off the tailing zero's; we don't need them.
   do; while(llGetSubString(mantissa, position, position) == "0" && (position = ~-position));//faster then a while loop
   
   //we do a bad thing, we recycle 'target' here, position is one less then target,
   //"target + ~position" is the same as "target - (position + 1)" saves 6 bytes.
   //this block of code actualy does the cutting.
   if(target + ~position) mantissa = llGetSubString(mantissa, 0, position);
   
   //insert the decimal point (not strictly needed). We add the extra zero for asthetics.
   //by adding in the decimal point, it simplifies some of the code.
   mantissa = llInsertString(mantissa, 1, llGetSubString(".0", 0, !position));

   //adjust exponent from having added the decimal place
   if((exponent += ~-target))
       mantissa += "e" + (string)exponent;
   //return with the correct sign.
   if(input < 0)
       return "-" + mantissa;
   return mantissa;

} </lsl>