How to convert decimal to Hex?



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192
4/18/2000 12:07:12 AM
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Usually it is customary to include body in your posts but here we go.

There are infinitely many ways to do this but the two I have used are
decimal to hex or decimal to binary to hex.  THe main advantage of the
latter is that it is extremely easy to go from binary to hex and from
decimal to binary.

First you need to know how many bytes are in the number you are dealing
with.  Each byte is two hexidecmal digits (digits isn't really the right
word but you know what I mean).  So for a two byte or 16 bit number there
are 4 hex digits.

hexidecimal digits go from 0 to 15 with fifteen = F  the number 16 decimal =
10 hex.  So to find each digit in the hex number you need to divide by 16 to
its position from the left minus one subtractr that number and do the next.
For example to turn 1000 into hex.

(% is the modulus operator in C)

1000 / 16^3 = 1000 / 4096 = 0

1000 % 16^3 = 1000

1000 / 16^2 = 1000 / 256 = 3

1000 % 256 = 232

232 / 16^1 = 232 / 16 = 14 = E

232 % 16 = 8

8 / 16^0 = 8 / 1 = 8

so the hex number is 03E8

This should be done in a loop but you're going to have to figure that one
out yourself.

192.168.0.2 <tech@chinachemfibre.com> wrote in message
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0
James
4/18/2000 12:30:23 PM
192,

to convert ll_decin to hex populate a string as follows:

//*******************
string ls_hexout, ls_decin
long ll_decin, ll_digit, ll_quotient

ls_decin = mle_1.text
ll_decin = long(ls_decin)
char lc_hexdigits[] = {'0','1','2','3','4','5','6','7', &
   '8','9','a','b','c','d','e','f'}

ll_digit = 7 // number of hex digits - 1
ls_hexout = "0x"
do while ll_digit >= 0
 ll_quotient = ll_decin / 16^ll_digit
 ll_decin = ll_decin - ll_quotient*(16^ll_digit)
 ls_hexout = ls_hexout + lc_hexdigits[ll_quotient + 1]
 ll_digit = ll_digit - 1
loop

Messagebox("hex equivalent of " + ls_decin, ls_hexout)
//*******************

Note that you can further optimize by keeping the results of "16^ll_digit"
in a local variable and only doing this calc once.
--
Regards,
Millard[Team Sybase]
Power3 - Custom Enterprise Training



192.168.0.2 wrote in message <5Q5klrMq$GA.286@forums.sybase.com>...
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0
Millard
4/18/2000 12:42:42 PM
Check out the numerical service from the PFC. Here's the code of the
of_hex(aul_decimal) function:

////////////////////////////////////////////////////////////////////////////
//
//
// Function:    of_Hex
//
// Access:    public
//
// Arguments:
// aul_decimal  Decimal value whose Hexadecimal representation needs to be
determined
//
// Returns:   string
//      The Hexadecimal representation of the decimal number.
//      If any argument's value is NULL, function returns NULL.
//
// Description:   Determines the Hexadecimal representation of a Decimal
number.
//
////////////////////////////////////////////////////////////////////////////
//
// Author: Lijun Yang  Lijun.Yang@worldnet.att.net
//
// Date:  4/7/97
//
//
// Revision History
//
// Date   Initials  Description of Change
// ----     --------  --------------------
// 4/7/97  L.Y.   Initial creation

string ls_hex=''
char lch_hex[0 to 15] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', &
       'a', 'b', 'c', 'd', 'e', 'f'}


If IsNull(aul_decimal) Then
 SetNull(ls_hex)
 Return ls_hex
End If

Do
 ls_hex = lch_hex[mod (aul_decimal, 16)] + ls_hex
 aul_decimal /= 16
Loop Until aul_decimal= 0

Return ls_hex

--

- Eric Aling [TeamSybase], AVALIX Information Systems, The Netherlands
  Eric's Home & PB Site @ http://www.knoware.nl/users/cypr115




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0
Eric
4/18/2000 1:47:51 PM
The following code converts decimal to any base:

function string uf_10toBase(integer ai_base, ulong al_number10):

string ls_rem, ls_num
integer li_rem

DO
    li_rem = Mod(al_number10,ai_base)
    if li_rem <= 9 then
        ls_rem = string(li_rem)
    else
        ls_rem = Char(55+li_rem)
    end if
    ls_num = ls_rem + ls_num
    al_number10 = al_number10 / ai_base
LOOP UNTIL al_number10 = 0
return ls_num

and to convert from any base to decimal:

function ulong uf_Baseto10(integer ai_base, string as_number):

integer   i, li_max
unsignedlong ll_return, ll_num
string   ls_char

as_number = Upper(as_number)
li_max = Len(as_number)
for i = li_max to 1 step -1
    ls_char = Mid(as_number,i,1)
    if IsNumber(ls_char) then
        ll_num = integer(ls_char)
    else
        ll_num = Asc(ls_char) - 55
    end if
    ll_num *= ai_base^(li_max-i)
    ll_return += ll_num
next
return ll_return

HTH

Simon



0
Simon
4/18/2000 2:41:04 PM
Reply:

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