I like to thank all of you for the information.
Cheers
Fawaz
--- Robert Colquhoun [EMAIL PROTECTED] wrote:
At 12:58 PM 5/03/2004, you wrote:
That is because UV (and I presumed UD) work on
two's complement integers, so
the top bit flips you between a positive and a
negative integer. This is why
I work the algorithm with 8 bit blocks, so that bit
1 in the 32 bit integer
is never 1 and so BITAND works as expected.
The problem was it was using signed 64 bit integers
everywhere except in
these functions which seem to use 32 bit integers,
as an example program on
unidata:
ININT=2147483648
CRT ININT = :ININT
CRT BITAND(ININT, 1) = :BITAND(ININT, 1)
CRT BITAND(ININT, 2^31 -1) = :BITAND(ININT,
2147483647)
CRT BITAND(ININT, 2 ^31) = :BITAND(ININT,
2147483648)
:TEST
ININT = 2147483648 // = 2 ^ 31
BITAND(ININT, 1) = 1// should be
0
BITAND(ININT, 2^31 -1) = 2147483647 // should be 0
BITAND(ININT, 2 ^31) = 2147483647 // should be
2^32
In your program you have BITAND(ININT, 255) to strip
out the smallest byte
which will work if and only if ININT 2^31
I probably should report this to ibm support, but
ibm outsourced our
support contract recently to a third party and i am
very nervous about
calling them up:
http://www.salon.com/tech/feature/2004/02/23/no_support/index_np.html
(Sorry about the ad, but article is one of the
funniest i have seen this year).
- Robert
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