>Dragan,
>Sergey passed this on to me. I need to do 4 byte floating point. You mention below that it can be done with some "tweaking". Well, what do I tweak?

The principle should be the same, but since you need 32 bits and not 64, it would have different sizes, and maybe a different offset for the exponent. I assume a there'd one byte of exponent and 3 bytes of mantissa. This function returns a numeric value for VFP's B format of 64-bit floating point number when passed as a string.

Since I didn't translate the variable names, here's their meaning:
bajt - byte
znak - sign
eksp(onent) - exponent
rez - res(ult)

I don't know the exact format of 32 bit floats - and I've seen a number of variations. The exponent may be offset by 128 (i.e. 128 would stand for zero, 129 for 1, 126 for -2 etc) or not; the bit used for the sign of the number itself may be in a different position etc. While working on this, it would help if you could write the numbers you want into strings (or a binary file) and then check whether your function returns proper values. Since I was doing this for Fox fields, it was no problem creating the file - I simply created a dbf with a B type field and filled it with numbers of my choice, repeating the same numbers in the next column as a string, so I could check my results.

>>This converts VFP's B format, which is 64-bit float. Tweaking it you can get the 32 bit version:
>>
>>

Function bton
>>*+ douBle string --> numeric
>>* takes a 8 byte string with IEEE representation of a long real
>>* structure:
>>* 7. byte:  7 bit sign
>>*	6-0 bit exponent upper
>>* 6. byte, bits 7-4 exponent lower. Exponent has 0x400 added
>>*    bits 3-0 mantissa. Mantissa is normalised, ie. assume 1 for first bit
>>* bytes 5-0 rest of mantissa
>>* returns a number.
>>*-
>>Lpara _s
>>Dime bajt(8)
>>For j=1 to 8
>>	bajt[j]=asc(subs(_s,j))
>>Endf
>>znak=iif(bittest(bajt[8], 7), -1, 1)
>>eksp=bitand(bajt[8],127)*16+bitrshift(bajt[7], 4)
>>*eksp=bitand(eksp+1025, 1023)
>>eksp=eksp-1023
>>mant=1+bitand(bajt[7], 15)/16
>>dv=4096
>>For j=6 to 1 step -1
>>	mant=mant +bajt(j)/dv
>>	dv=dv*256
>>Endf
>>rez=mant*2**eksp*znak
>>Retu rez