# convert HEX to double

Discussion in 'Java' started by puzey@yahoo.com, Mar 29, 2006.

1. ### Guest

Hi there,

Need your expertise here. I am trying to convert a Hex in string format
to double. e.g.

String toConvert = "bfdf000000000000";
Double dblDouble = Double.longBitsToDouble(Long.parseLong(toConvert,
16));
System.out.print(dblDouble);

I got a NumberFormatException because bfdf000000000000 is out of range
for a long type. But this number is within the range of double. How can
I convert?

Puze

, Mar 29, 2006

2. ### Thomas SchodtGuest

wrote:
> Hi there,
>
> Need your expertise here. I am trying to convert a Hex in string format
> to double. e.g.
>
> String toConvert = "bfdf000000000000";
> Double dblDouble = Double.longBitsToDouble(Long.parseLong(toConvert,
> 16));
> System.out.print(dblDouble);
>
> I got a NumberFormatException because bfdf000000000000 is out of range
> for a long type. But this number is within the range of double. How can
> I convert?
>
>
> Puze
>

http://java.sun.com/j2se/1.4.2/docs/api/java/math/BigInteger.html#BigInteger(java.lang.String, int)

http://java.sun.com/j2se/1.4.2/docs/api/java/math/BigInteger.html#doubleValue()

Thomas Schodt, Mar 29, 2006

3. ### Thomas WeidenfellerGuest

wrote:
> I got a NumberFormatException because bfdf000000000000 is out of range
> for a long type. But this number is within the range of double. How can
> I convert?

For example by just coding the conversion by hand:

long result = 0;
String s = "fbcd000000000000"
for(int i = 0; i < s.length(); i++) {
result = (result << 4) | Character.digit(s.charAt(i), 16);
}

You need to add error handling if s can be to long or s not necessarily
includes hex digits only.

/Thomas
--
The comp.lang.java.gui FAQ:
http://www.uni-giessen.de/faq/archiv/computer-lang.java.gui.faq/

Thomas Weidenfeller, Mar 29, 2006
4. ### Guest

Thanks, Thomas Weidenfeller! It seems work.

Thanks also go to Thomas Schodt and others who send me email directly.

Have a nice day!

Puze

, Mar 29, 2006
5. ### Roedy GreenGuest

On 29 Mar 2006 06:16:36 -0800, wrote, quoted or
indirectly quoted someone who said :

>String toConvert = "bfdf000000000000";
>Double dblDouble = Double.longBitsToDouble(Long.parseLong(toConvert,
>16));
>System.out.print(dblDouble);

The problem is parseLong is expecting a signed number. The unsigned
one you give it is too big.

You can see the problem if you run this code fragment
String toConvert = "bfdf000000000000";
long temp = Long.parseLong( toConvert, 16 );
System.out.println( temp );

Your code contains another error, masked by boxing. This is closer to
what you meant:

long temp = 0xbfdf000000000000L;
double d = Double.longBitsToDouble( temp );
System.out.println( d );

Double.longBitsToDouble returns a double, not a Double.

I see three ways to solve your unsigned problem:

1. arrange to get get the bits in binary rather than hex.

2. construct the low and high 32 bits then shift then OR.

3. Convert hex chars from unsigned to signed, perhaps when generated
(see http://mindprod.com/jgloss/hex.html) or by checking for high
char being >= 8 , replace it with equivalent char <8 with high bit
off, then mask the sign bit back in afterwards. Conceptually simple,
but in practice long winded.

For hints on how to do the details, so the at the code in
http://mindprod.com/products1.html#LEDATASTREAM