I think it treats the hex value as positive as it can be preceeded by
a minus sign. From the docs:
The string may be an optional minus sign - followed by a
hexadecimal (0x... or #...), octal (0...), or decimal (...)
representation of a long.
I agree it is unclear.
On Aug 10, 8:29 pm, RLScott
On Wed, Aug 10, 2011 at 4:48 PM, RichardC richard.crit...@googlemail.comwrote:
The string may be an optional minus sign - followed by a
hexadecimal (0x... or #...), octal (0...), or decimal (...)
representation of a long.
I agree it is unclear.
It's not unclear at all This sequence of
In Java, integer types are signed.
10.08.2011 23:48, RichardC пишет:
I think it treats the hex value as positive as it can be preceeded by
a minus sign. From the docs:
The string may be an optional minus sign - followed by a
hexadecimal (0x... or #...), octal (0...), or decimal (...)
The problem is that 0x9774d56d682e549c is a positive value that's too
large to fit in a 64 bit long. If the intent is that the high bit of
the first digit indicates sign, then you'll need to mask off the high
bit and add a leading minus sign when the high bit is on. E.g.
On Aug 10, 2:50 pm, Nick Risaro nris...@gmail.com wrote:
It's not unclear at all This sequence of characters *must represent a
positive value* or a
NumberFormatExceptionhttp://download.oracle.com/javase/6/docs/api/java/lang/NumberFormatEx...will
be thrown
The docs for the Long class say
Well, it is no longer an issue. I decided to work with the byte array
(String.getBytes(UTF-8)) instead.
--
You received this message because you are subscribed to the Google
Groups Android Developers group.
To post to this group, send email to android-developers@googlegroups.com
To unsubscribe
6 matches
Mail list logo
