How to conver 4 bytes to double floating point?

[Natalie Angel]

Hi,

I got a new question to you all

If I have 8 single bytes in HEX like 00 00 00 40 A6 54 58 41 how can I convert this to double floating point which mean I should have an answer which looks like this; 6378137.000000. I thought the max on both sides of the "." should be 65535 4 bytes but I can see that is not the case?

I did try the shown sample but the answer is wrong as I test the reply from another software that works;

Val("&H" & Mid$(sGTData, 45, 8)) & "." & Val("&H" & Mid$(sGTData, 53, 8))

where sGTData is a long HEX string and I then read from position 45 4 bytes and position 53 4 bytes (in HEX).

The answer should be 6378137.000000 but I get 64.-1504421823 where the first value is 00000040 and the second A6545841 in HEX. I can se that HEX 40 is 64 but how can the answer be 6378137.000000? Any ideas out there?

[Michael Mattias]

TRY:

UNION Foo 
   D  AS DOUBLE
   B (1 TO 8) AS BYTE
END UNION

LOCAL DU AS Foo 
  FOR Z = 1 TO 8 
      DU.B(Z) =  VAL ("&H" & MID$(stringin,(Z-1)*2 + 1, 2))
   NEXT
   PRINT FORMAT$(DU.D)

If that gives you bad answer, try it reversing the bytes of the input string.

[John Gleason]

Here is one possibility:

#COMPILE EXE
#DIM ALL

FUNCTION PBMAIN () AS LONG
   LOCAL d AS DOUBLE
   LOCAL sRev AS STRING
   
   sRev = STRREVERSE$(MKQ$(&h00000040A6545841)) 'reverse the byte order 1st
   d = CVD(sRev)                                'convert it to a double
   ? FORMAT$(d, "0.000000")                     'display it
   
END FUNCTION

[Natalie Angel]

Thank you both.

I was using
Sub fnConvertReverse32HEX(sHexString As String)
sNewData = ""
For iLoop = 7 To 1 Step -2
sTmp = Mid$(sHexString, iLoop, 2)
sNewData = sNewData & sTmp
Next iLoop
End Sub

But I will try John's sample